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\begin{document}
\title{Introduction to Category Theory}




\address{ $\dagger$ Institut für Mathematik, Universität Zürich, Winterthurerstrasse 190 CH-8057 Zürich}



%\begin{abstract}
%\end{abstract}


\maketitle

\tableofcontents

%%--------------------------------------
\section{Instructions}
Welcome one and all to our shared exercise sheet! The purpose of this space is two-fold. Firstly, because the only way to learn is to do, and listening to your peer's talks is not enough to understand category theory. Secondly, because our goal is vast, and if we were to cover the whole book (plus our extra topics at the end) we would need twice as much time. 

In class the speaker will always explain the most important topics, but you should always try, before or after the lecture, to read the relevant chapter on your own and attempt one of the questions below. These will allow you a chance to learn, and the speaker an excuse to skip the less interesting topics.
\hfill \break

\begin{itemize}
\item Questions, answers and comments will be posted the following way: 
\end{itemize}


\subsection{} What is the most creative color?\hfill \break

\iam{Student}
I don't know but it is not green.

\iam{An Other Student}
YELLOW!

\subsection{} The next Question...\hfill \break


\begin{itemize}
\item Remember to sign you name using the \textbackslash  iam\{\}  command. Try to answer these questions without fear, ask your fellow students about their answers, or even post questions of your own!

\item If you would like to draw simple diagram you can do so here https://tikzcd.yichuanshen.de and insert them in the tex file as follows (see code)

\begin{center}
% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAEEQBfU9TXfIRQAmclVqMWbAELdeIDNjwEio4ePrNWiEAGFu4mFADm8IqABmAJwgBbJGRA4ISAIzVNUnRbmWb9xHcnF0RRCS02Y18QazsHamckMM9tEGMAHXSAYywrLIACH2oGOgAjGAYABX5lIRArLGMACxwDLiA
\begin{tikzcd}
A \arrow[rr, "f"] \arrow[rrdd, "g\circ f"'] &amp;  &amp; B \arrow[dd, "g"] \\
                                            &amp;  &amp;                   \\
                                            &amp;  &amp; C                
\end{tikzcd}
\end{center}
\end{itemize}
\section{Week 1}

\subsection{} \label{Rel} The category \textbf{Rel} has as objects sets and morphisms $f:A\to B$ are \emph{relations}, that is subsets of the form $f\subseteq A\times B$. Composition in this category is given by
\begin{align*}
g\circ f = \{ (a,c)\in A\times C \,|\, \exists b\in B \,\,\text{with}\,\, (a,b)\in f, (b,c)\in g\}
\end{align*}
for $f\in A\times B$ and $g\in B\times C$. Show that \textbf{Rel} satisfies the axioms of a category. What is the identity morphism?\hfill \break

\iam{Sophie} 
\begin{enumerate}
\item Associativity: Let A, B, C, D be sets and $f:A\to B$, $g:B\to C$  $h:C\to D$
Then we have 
\begin{align*}
h\circ (g \circ f) &amp;= \{(a,d)\in A\times D \,|\, \exists c\in C \,\,\text{with}\,\, (a,c) \in g\circ f, (c,d) \in h\}
\\
&amp;=\{(a,d)\in A\times D \,|\, \exists b\in B, \, \exists c\in C\,\text{with}\,\, (a,b) \in f, (b,c) \in g, (c,d) \in h\}
\\
&amp;= \{(a,d)\in A\times D \,|\, \exists b\in B \,\,\text{with}\,\, (a,b) \in f, (b,d) \in h\circ g\}
\\
&amp;=(h\circ g)\circ f 
\end{align*}
\item Identity Morphism: For any set A, let $\id_A := \{(a,b) \in A\times A \,| \ a = b\} $ . Then for any morphism $f:A\to B$, if (a,b) $\in$ f then (a,b) $\in f\circ \id_A$ , using (a,a) $\in \id_A$. Conversely if (a,b) $\in f\circ \id_A$, then (a,a) $\in \id_A$ and thus (a,b) $\in f$. This shows $f=f\circ \id_A$. In the same way we can show that for any morphism $g:C\to A$ we have that $g=\id_A \circ g$.
\end{enumerate}
=======\\
\iam{Olga}
According to p.5 of Awodey's book we have to show that associativity and identity laws hold true:\\

1. Associativity:\\
For any $f : A \rightarrow B$, $f : B \rightarrow C$, $f : C \rightarrow D$ morphisms in the category \textbf{Rel}:\\
\begin{align*}
h\circ (g\circ f ) = h \circ \{(a,c)\in A \times C \,|\, \exists \,b  \in B \, \text with \, (a,b)\in f, (b,c)\in g \} \\
= \{(a,d)\in A \times D \,|\, \exists \,c  \in C \, \text with \, (a,c)\in g\circ f, (c,d)\in h \, \\
\text{and} \, \exists \, b  \in B \, \text with \, (a,b)\in f, (b,c)\in g \}\\
= \{(a,d)\in A \times D \,|\, \exists \, b  \in B \, \text with \, (a,b)\in f, (b,d)\in h\circ g \\
\text{and} \, \exists \, c  \in C \, \text with \, (b,c)\in g, (c,d)\in h \}\\
=(h\circ g) \circ f
\end{align*}\\
2. Unit: \\
We have the following identity morphism:
\begin{align*}
1_A = \{ (a,a) \, \in A\times A \, | a\in A\} \subseteq A\times A
\end{align*}
Let $f : A \rightarrow B$ be a binary relation, then:
\begin{align*}
f\circ 1_A = \{ (a,b) \in A\times B \, | \, \exists \, a\in A \, \text{with} \, (a,a)\in 1_A , (a,b) \in f\} = f = \{ (a,b) \in A\times B \, | \, \exists \, b\in B \, \text{with} (a,b) \in f, (b,b)\in 1_B \}  = 1_B\circ f 
\end{align*}
where we have used the fact that identity arrow exists for each object A and B.\hfill \break


\subsection{} For a fixed set $X$, we call $\mathcal{P}(X)$ its \emph{power set}. This is the set of all subsets of $X$. Show how this has naturally the structure of a poset (so in particular a simple category) using subset inclusions.\hfill \break

\iam{Arno} We have to check three properties, reflexivity, transitivity and antisymetry. \\
1. Reflexivity: Clearly this holds, since every set is contained in itself.\\
2. Transitivity: Checking this on elements gives us this result.\\
3. Antisymmetry: If two sets are simultainiously contained in one another, they have to have the exact same elements, hence are equal as sets.

\subsection{}  Check that functors compose in an associative way, proving that $\mathbf{Cat}$ is a category. \hfill \break

\iam{Muhammed}  Let $\mathbf{C_1}, \mathbf{C_2}, \mathbf{C_3}, \mathbf{C_4}$ be  categories and $F: \mathbf{C_1} \to \mathbf{C_2}, \, G: \mathbf{C_2} \to \mathbf{C_3}, \, H: \mathbf{C_3} \to \mathbf{C_4}$ functors. We consider 
$$K:= G \circ F:  \mathbf{C_1} \to  \mathbf{C_3},$$
where for any object $A$ and any arrow $f$ in $\mathbf{C_1}$, $$
K(A) := G(F(A))  \quad \text{and} \quad K(f) := G(F(f)) \quad \quad (2.3.1)$$
We show that $K$ is a functor. \\ Let $A,B$ be objects and $f: A \to B, \, g: B \to C$ arrows in $\mathbf{C_1}$. \begin{enumerate}
\item Since $F,G$ are functors, $F(f): F(A) \to F(B)$ and $G(F(f)): G(F(A)) \to G(F(B))$. Thus, $K(f): (G \circ F)(A) \to (G \circ F)(B)$.
\item $(G \circ F)(f \circ g) := G(F(f \circ g)) = G(F(f) \circ F(g))= G(F(f)) \circ G(F(g))$.
\item $K(1_A) := G(F(1_A)) = G(1_{F(A)}) = 1_{G(F(A))} =: 1_{(G \circ F)(A)}$
\end{enumerate}
Also, clearly, $K(A)$ is an object in $\mathbf{C_3}$. This proves the assertion that $K$ is a functor.\\
To show associativity of the composition, we require
$$
H \circ (G \circ F) = (H \circ G) \circ F \quad (2.3.2) 
$$ 
where both sides are functors from $\mathbf{C_1}$ to $\mathbf{C_4}$ by previous considerations.
Now, by (2.3.1), it readily follows that, 
$$
(H \circ (G \circ F))(A) = H((G \circ F)(A)) = H(G(F(A))) = (H \circ G)(F(A)) = ((H \circ G) \circ F)(A) 
$$
and
$$(H \circ (G \circ F))(f) = H((G \circ F)(f)) = H(G(F(f))) = (H \circ G)(F(f)) = ((H \circ G) \circ F)(f)$$
for every any object $A$ and any arrow $f$ in $\mathbf{C_1}$. This yields the desired equality stated in (2.3.2).


\subsection{}  Prove that inverses are unique. \hfill \break

\iam{Rızacan} Let $f$ be an arrow with Cod($f$)=$B$ and Dom($f$)=$A$. Suppose
$g,g'$ are both inverses of $f$. Then by associativity and properties of
the identity: $g'=\id_A\circ
g'=(g\circ f)\circ g'= g\circ(f\circ g')=g\circ \id_B=g$.


z\subsection{} Argue whether the following \emph{isomorphisms} of categories hold:
\begin{enumerate}
\item $\mathbf{Rel}\cong \mathbf{Rel}^{op}$;
\item $\mathbf{Set}\cong \mathbf{Set}^{op}$;
\item For a fixed set $X$, $\mathcal{P}(X)\cong \mathcal{P}(X)^{op}$.
\end{enumerate}\hfill \break

\subsection{} Show that in $\mathbf{Set}$ isomorphisms are bijections.\hfill \break

\iam{Raphael} Let $f\in \mathbf{Set}(X,Y)$ be an isomorphism of sets, i.e. a morphism so that there is another morphism $g:Y\rightarrow X$ with
$g\circ f=\id_X$ and $f\circ g=\id_Y$.\\
Injectivity: Let $x_1,x_2\in X$ with $f(x_1)=f(x_2)$, then composing $g$ to the left implies immediately $x_1=x_2$ as a direct consequence of the definition of an isomorphism.\\
Surjectivity: Let $y\in Y$ be arbitrary and define $x=g(y)$, then $f(x)=f(g(y))=\id_Y(y)=y$.\\
This shows, that $f$ is indeed a bijection.

\subsection{} Show that in $\mathbf{Mon}$, the category of monoids, isomorphisms are bijective homomorphisms.\hfill \break

\iam{Andreas} Updated: Let $N,M$ be two monoids and $f:N\to M$ be a monoid homomorphismus. Suppose $f$ is a bijective homomorphismus, this means there exist an $n\in N$ such that $f(n)=m$ for every $m\in M$ by surjectivity and since $f$ is injective this $n$ is unique and we can define $g:M\to N$ as $g(m)=n$. So we have $g(f(n))=n$ respectively $f(g(m))=m$ and thus $g\circ f=1_N$ and $f\circ g=1_M$. We try to show that $g$ is a monoid homomorphismus too: let $m_1,m_2\in M$, then
we have $$f(g(m_1)g(m_2))=f(g(m_1))f(g(m_2))=m_1m_2=f(g(m_1m_2))$$ Since we assume $f$ is bijective it follows that
$$g(m_1m_2)=g(m_1)g(m_2)$$
and thus a group homomorphism, hence a bijective homomorphism is an isomorphism in the category $\textbf{Mon}$.

\iam{Nicola} I'm not convinced by your logical argument. You assume that $f$ is a bijective homomorphism, then in the next sentence you follow with the existence of $g$. But this is true if you assume that $f$ is an iso (by definition), which is what you want to prove!

 Perhaps this argument is better stated as ``...\emph{then there exists a map of sets $U(g):U(M)\to U(N)$ such that ...}'', where $U:\mathbf{Mon}\to \mathbf{Set}$ is the functor that forgets about the multiplication. This is a small point, but for these types of arguments is important to value precision. 
 
 
\subsection{} Show that in $\mathbf{Pos}$, the category of posets, isomorphisms are \emph{not} bijective homomorphisms.\hfill \break

\iam{Marius} Consider the posets $\mathcal{P} := 
	\begin{tikzcd}[cramped, sep=scriptsize]
	a \arrow[loop, distance=1.2em, in=115, out=65] &amp; b \arrow[loop, distance=1.2em, in=115, out=65]
	\end{tikzcd}$ and $\mathcal{Q} := 
	\begin{tikzcd}[cramped, sep=scriptsize]
	a' \arrow[loop, distance=1.2em, in=115, out=65] \arrow[r] &amp; b' \arrow[loop, distance=1.2em, in=115, out=65]
	\end{tikzcd}$, along with the map $f: \mathcal{P} \rightarrow \mathcal{Q}$ sending $a \mapsto a'$, and $b \mapsto b'$. Clearly, $f$ is bijective and monotone. However, it has no inverse: There are exactly two monotone maps from $\mathcal{Q}$ to $\mathcal{P}$, one sending both $a',b' \mapsto a$, and another sending both $a',b' \mapsto b$. Since neither is surjective, we can't recover the identity on $\mathcal{P}$ by pre-composing with $f$. Hence, $f$ is a bijective homomorphism which is not an isomorphism, as desired.

\subsection{} Construct the co-slice category $A\downarrow \mathbf{C}$, of objects of $\mathbf{C}$ ``under'' $A$, using the slice category $\mathbf{C}\downarrow A$ and the operation $(-)^{op}$. Prove your claim.\hfill \break

\subsection{} How many \emph{free categories} are there with 6 arrows? \emph{Hint:} Try to draw them on a piece of paper, and if you feel industrious you can report them here using this online tool https://tikzcd.yichuanshen.de/ \hfill \break

\iam{Jessica} Consider quivers with at most 6 vertices (otherwise there would be already more than 6 identity arrows). Quivers with edges such that loops become valid paths are not considered, since it would have an arrow for every finite path involving the loop. 

To see an example of this, let $Q$ be the quiver with two vertices $a, b$ and two edges $a \stackrel{f}{\rightarrow} b,\ b \stackrel{g}{\rightarrow} a$. The free category on $Q$ then has two identity arrows on the vertices and an arrow for every finite sequence involving the loop, including $f,\ g,\ g \circ f,\ f \circ g,\ f \circ g \circ f,\ g \circ f \circ g$ etc. 

Keeping the above in mind, we have below the drawings of all possible quivers (without the identity paths) such that the free categories on them have precisely 6 arrows. 

\textbf{\# vertex = 1}. Since loops are not allowed, the quiver with only one vertex is not considered. 

\textbf{\# vertices = 2}. Only one quiver is possible. On top of the two identity arrows, the free category on this has four arrows from one vertex to the other vertex like so:
\[
	\begin{tikzcd}
	A \arrow[r, shift left] \arrow[r, bend left] \arrow[r, shift right] \arrow[r, bend right] &amp; B
	\end{tikzcd}
\]

\textbf{\# vertices = 3}. Four quivers are possible. On top of the three identity arrows, the free categories on these have exactly three more possible arrows as shown in the diagrams below (the third arrow of the first quiver is $g \circ f$). 
\[
	\begin{tikzcd}
	A \arrow[r, "f"] &amp; B \arrow[d, "g"] &amp;  &amp; \
	A \arrow[rd] \arrow[rd, bend right] \arrow[r] &amp; B &amp;  &amp; \
	A \arrow[rd] \arrow[rd, bend right] &amp; B \arrow[d] &amp;  &amp; \
	A \arrow[r] \arrow[r, bend right] \arrow[r, bend left] &amp; B \ \\
	&amp; C   &amp;  &amp;   &amp; C   &amp;  &amp;   &amp; C   &amp;  &amp;   &amp; C
	\end{tikzcd}
\]

\textbf{\# vertices = 4}. Four quivers are possible. On top of the four identity arrows, the free categories on these exactly two more possible arrows as shown below:
\[
	\begin{tikzcd}
	A \arrow[r, shift left] \arrow[d] &amp; B &amp;  &amp; A \arrow[r] &amp; B &amp;  &amp; A \arrow[r]  &amp; B &amp;  &amp; A \arrow[r, shift right] \arrow[r, shift left] &amp; B \\
	D                                 &amp; C &amp;  &amp; D \arrow[r] &amp; C &amp;  &amp; D \arrow[ru] &amp; C &amp;  &amp; D                                              &amp; C
	\end{tikzcd}
\]

\textbf{\# vertices = 5}. Only one quiver is possible. On top of the five identity arrows, the free category on this has one arrow from a vertex to another vertex like so: 
\[
	\begin{tikzcd}
	A \arrow[r] &amp; B \\
	D           &amp; C  &amp; E
	\end{tikzcd}
\]

\textbf{\# vertices = 6}. Only one quiver is possible, and it has no arrows aside from the six identity arrows. 

\subsection{} (\emph{Harder}) Prove the universal mapping property of free categories on graphs. \hfill \break

\section{Discussion: Week 1}

\iam{Severin} I find it useful to also allow for some informal discussion in this file. Nicola also agreed on that. So if there is a statement that puzzles you, it probably will be also confusing for the other people.

\iam{Severin} IN Awodey, p. 22 it says "For example, all finite categories are clearly small, as is the category $\textbf{Sets}_{\text{fin}}$ of finite sets and functions." This confused me a bit, as it is not true. Indeed, we can consider the class (this is the generalized notion for stuff that is too big to be a set, just like the collection of all sets)
$$ \{ \{M\} \ : \ M\in \textbf{Set} \}. $$
Nicola explained me that the statement is at least morally true. The category is "roughly the same" as some category that is small. The main problem is that we allow for too many copies of the same thing. There is no big difference between $\{\lozenge\}$ and $\{ \heartsuit\}$ (as sets). The correct notion of "roughly the same" will be equivalence of categories, which we can find in Chapter $7$ in Awodey. In fact our example is explicitely treated in Example $7.23.$ (p. $146$). 

\iam{Marius} Regarding, the above, I'm not sure your counterexample works. If I understand correctly, you are saying that the collection $\Theta := \{ \Psi \}$, where $\Psi := \{M : M \in \mathbf{Set}\}$, is finite, since it contains only the single element $\Psi$ (I'm unclear why you put $\{M\}$ in your example?). On the other hand since, $\Theta$ is a proper class (if it where a set, $\bigcup{\Theta} = \Psi$ would also be by the axiom of union, contradiction by Russell's paradox). In particular, the collection $\Theta$ is not a set and the category $\mathcal{C}$ with $Ob(\mathcal{C}) := \Theta$ is not small. 

Assuming this is the argument, I doubt that $\Theta$ finite. In set theory we define a set to be finite if it is in bijection with an element of the natural numbers (as constructed set-theoretically from ordinals and the axiom of infinity). Since $\Theta$ is a proper class, there aren't any set-theoretic mappings between $\Theta$ and any set: Any such mapping is a subset of the cartesian product of two sets. Hence, if we assume such as map, we could again union our way to a set of all sets. Therefore, it seems we would need an extension of the concept of finiteness if we wanted to apply it to proper classes. Maybe, we could re-construct an analogue of the natural numbers at each level of the class hierarchy and then count elements with respect to these? Under the set-theoretic definition, however, $\Theta$ fails to be finite and is therefore not a counterexample.

\iam{Severin} I am not sure, whether I understand, what you mean. I want to show that the collection of all finite set is a proper class (and thus $\textbf{Set}_{\text{fin}}$ is not a small category). For this I note that the collection of all singeltons is a proper class. What I am saying is that $\{M\}$ is finite (clearly $M$ is not finite in general and thus I wrote the brackets). I am not sure, why you introduce $\Psi$. Can you clarify?

\iam{Marius} Okay, I see now what you were objecting to the claim that $\textbf{Set}_{\text{fin}}$ is a small category. I now agree with you. Thanks for pointing that out! I implicitly assumed we could somehow enumerate finite sets based on cardinality, but this is clearly incorrect upon reflection, as your example shows. Because of this mistaken assumption I thought you were objecting to the claim that all finite categories are small. I apologize for any confusion I created. I believe my example still shows that a category might not be small, even if it only contains a single object.

What you report Nicola as saying makes more sense now. Indeed, we can consider the category whose objects are isomorphism classes of finite sets. Restricting to some choice of representatives, we obtain an equivalent subcategory of $\textbf{Set}_{\text{fin}}$ which contains no isomorphic objects (called the skeleton of $\textbf{Set}_{\text{fin}}$, see e.g. wikipedia). This skeleton will then be isomorphic as a category to $\mathbf{FinOrd}$, the category of finite ordinals, a small category.

\iam{Severin} Indeed, that is the example I referenced in the book. I agree that a category might not be small, even if it contains only one element (unless that element happens to be a set). Something that I find very surprising!

\iam{Muhammed} I am not sure if I understand the charactarization of concrete categories through test objects as described in Rmk. 1.7? Can somebody elucidate this?

\section{Week 2}



\subsection{} Show that, in $\mathbf{Set}$, epis are surjective functions. Conclude that isos in this category are morphisms that are epic and monic. Give an example of a category where this is not the case.\hfill \break
\iam{Severin} One example where it does not hold is any given poset (with more than two points that are comparable and distinct) viewed as a skeletal category (all arrows are monic and epic by $4.2$, but the arrow between the two comparable distinct points is not an iso as there is no arrow in the opposite direction to begin with). Another example we saw in Olga's talk, the category of monoids (with the explicit example given by the inclusion map from $(\mathbb{N}, +, 0)$ to $(\mathbb{Z}, +, 0)$). We can tweak this example a bit and get an example in the category of rings (objects are rings and arrows are ring homomorphisms). Namely, we can take the inclusion from $(\mathbb{Z}, +, \cdot)$ to $(\mathbb{Q}, +, \cdot)$. More generally, we can take integral domain $(R, +, \cdot)$ that is not a field and consider the "inclusion" into its quotient field $\operatorname{Quot}(R)$. Another example would be the category of smooth manifolds (objects are smooth manifolds and arrows are smooth maps). Then the arrow $f: \mathbb{R}\rightarrow \mathbb{R}, \ f(x)=x^3$ yields an arrow that is both epic and monic, but not an iso. Feel free to show that those are really examples of arrow that are both epic and mono, but not iso.
\subsection{} Show that in a poset, treated as a skeletal category, all arrows are both monic and epic.\hfill \break
\iam{Severin} In a poset there exists at most one arrow between two given objects. Thus, already saying $g,h$ are arrows between $B$ and $D$ implies that $g=h$ (without even mentioning $f$).


\subsection{} For the following exercises, consider the typical commutative diagram 
\begin{center}\begin{tikzcd}
A \arrow[rr, "f"] \arrow[rrdd, "g\circ f"'] &amp;  &amp; B \arrow[dd, "g"] \\
                                            &amp;  &amp;                   \\
                                            &amp;  &amp; C                
\end{tikzcd}
\end{center}
in any category $\mathbf{C}$. 
\begin{enumerate}
\item Show that if $f$ and $g$ are isos (resp. monos or epis), then so is $g\circ f$;
\item Show that if $g\circ f$ is monic, then so is $f$;
\item Show that if $g\circ f$ is epic, then so is $g$;
\item Give an example of $g\circ f$ being monic (or resp. epic) while $g$ (or resp. $f$) is not.
\end{enumerate}
\hfill \break

\iam{Olga}
\begin{enumerate}
\item \underline{$f,g$ are isos $\implies (g \circ f)$ is iso}
\begin{proof}
Assume f, g are isos \\then $\exists f^{-1}, s.t. f^{-1} \circ f= 1_A$ and $f \circ f^{-1}=1_B $\\ and $\exists g^{-1}$, s.t. $g^{-1} \circ g= 1_B$ and $g \circ g^{-1}=1_C $ \\
We want to show that for $(g \circ f): A \to C$,  $\exists (g \circ f)^{-1}$ s.t. $(g \circ f)^{-1} \circ (g \circ f) = 1_A$ and $(g \circ f) \circ (g \circ f)^{-1}= 1_C$\\
So we have:\\
$(g \circ f)^{-1} \circ (g \circ f) = (f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1} \circ g) \circ f = f^{-1} \circ 1_B \circ f = f^{-1} \circ f = 1_A$ \\
$(g \circ f) \circ (g \circ f)^{-1} = (g \circ f) \circ (f^{-1} \circ g^{-1}) = g \circ (f \circ f^{-1}) \circ g^{-1} = g \circ 1_B \circ g^{-1} = g \circ g^{-1} = 1_C$\\
$\implies (g \circ f)$ is iso.\\
Moreover we can show that \underline{if $f, g$ are monos then so is $(g \circ f)$}:
\begin{proof}
Assume $f$ is mono, i.e. for some $h, \tilde{h}:D \to A$ and $k, \tilde{k}: B \to E$:\\
if $f \circ h = f \circ \tilde{h}$ then $h = \tilde{h}$\\
if $k \circ f = \tilde{k} \circ f$ then $k = \tilde{k}$\\
Analog for $g$.\\

Take $h, \tilde{h}:D \to A$ and consider: \\
$(g \circ f) \circ h = (g \circ f) \circ \tilde{h}$\\
since composition is assoc. $g \circ (f \circ h) = g \circ (f \circ \tilde{h})$\\
since g is mono $f \circ h = f \circ \tilde{h}$\\
since f is mono $h = \tilde{h}$\\
$\implies (g \circ f)$ is mono\\

Now take $k,\tilde{k}:C \to F$ and consider: \\
$k \circ (g \circ f) = \tilde{k} \circ(g \circ f)$ \\
since composition is assoc. $(k \circ g) \circ f = (\tilde{k} \circ g) \circ f$\\
since f is epi $k \circ g = \tilde{k} \circ g$\\
since g is epi $k = \tilde{k}$\\
$\implies (g \circ f)$ is epi
\end{proof}
\end{proof}
\item \underline{$g \circ f$ is mono $\implies f$ is mono}
\begin{proof}
We want to show that if $f \circ h = f \circ \tilde{h}$, then $h = \tilde{h}$\\
Consider $f \circ h = f \circ \tilde{h}$ for some $h, \tilde{h}:D \to A$\\
combine with g on left $g \circ f \circ h = g \circ f \circ \tilde{h}$ \\
because of assoc. $(g \circ f) \circ h = (g \circ f) \circ \tilde{h}$ \\
since $g \circ f$ is mono $h = \tilde{h}$ \\
$\implies f$ is mono
\end{proof}
\item \underline{$g \circ f$ is epi $\implies g$ is epi}
\begin{proof}
We want to show that if $k \circ g = \tilde{k} \circ g$, then $k = \tilde{k}$\\
Consider $k \circ g = \tilde{k} \circ g$ for some $k,\tilde{k}:C \to F$\\
combine with f on right $k \circ g \circ f = \tilde{k} \circ g \circ f$ \\
because of assoc. $k \circ (g \circ f) = \tilde{k} \circ (g \circ f)$ \\
since $g \circ f$ is epi $k = \tilde{k}$ \\
$\implies g$ is epi
\end{proof}
\item Example\\
Consider category \underline{Set} and following functions\\
$f: [0, \inf) \to \mathbb{R}$, $f(x)=x$ is injective \\
$g: \mathbb{R} \to [0, \inf)$, $g(x)=|x|$ is not injective (since $g(-1)=g(1)$, but $1\neq{-1}$)\\
and \\
$(g \circ f) : [0, \inf) \to [0, \inf)$, $(g \circ f)(x)=|x|$ is bijective, i.e. injective
\end{enumerate}

\subsection{} $\mathbf{Vec}_k$ is the category of vector spaces over the field $k$ and linear maps between them. Describe its initial and terminal object. \hfill \break

\iam{Muhammed} The initial  and terminal object in $\mathbf{Vec_k}$ is the trivial vector space $W = \{0_W\}$. Indeed, let $V$ be an object in $\mathbf{Vec_k}$ Then, the unique linear map from ${0_W}$ to $V$ is given by $w \mapsto 0_V$. Similarly, the unique homomorphism from $V$ to $\{0_W\}$ is characterized by the mapping $v \mapsto 0_W$. As these are true for any arbitrary $V$ in the category, the claim follows. 


\subsection{} Show that the following are equivalent. For an arrow $f:A\to B$ in any category,
\begin{enumerate}
\item $f$ is an iso;
\item $f$ is both a mono and a split epi;
\item $f$ is both epic and a spit mono;
\item $f$ is both a split mono and a split epi.
\end{enumerate}\hfill \break
\iam{Andreas}
\begin{itemize}
\item $(i)\to (iv)$: $f$ is an isomorphism, meaning we have $ff^{-1}=1_B$ and $f^{-1}f=1_A$, so we have
$f(f^{-1}f)=f\circ 1_A$ and $(ff^{-1})f=1_B\circ f$, hence there exist right resp. left inverse, thus $f$ is a split epi and split mono.
\item $(iv)\to (i)$: If $f$ is both split mono and epi, there exist a left and right inverse such that $f^{-1}f = 1_A$ and $ff^{-1}=1_B$, but that is the definition of an isomorphism.
\item $(i)\to (ii)$: By $(i\to iv)$ we know that $f$ is split epi. Consider arrows $i,j$ such that $fi=fj$, then it follows that $(f^{-1}f)i=(f^{-1}f)j$ and thus $i=j$, hence $f$ is a mono.
\item $(i)\to (iii)$: By $(i\to iv)$ we know that $f$ is split mono. Consider arrows $i,j$ such that $if=jf$, then it follows that $i(ff^{-1})=j(ff^{-1})$ and thus $i=j$, hence $f$ is a epic.
\item $(ii)\to (iv)$ So $f$ is a mono, so for every arrow $i$ with $fi=ff^{-1}$ it follows that $i=f^{-1}$. Since $f$ is split epi, there exists such a right inverse $f^{-1}$.
\item $(iii)\to (iv)$ So $f$ is a epic, so for every $j,i$ with $if=f^{-1}f$ it follows that $i=f^{-1}$. Since $f$ is split mono, there exists such a left inverse $f^{-1}$.
\end{itemize}

\subsection{} Show that in any category, any retract of a projective object is projective. \hfill \break

\iam{Muhammed} Let $\mathbf{C}$ be a category, $P$ projective and $A$ retract of $P$. Then, there exist arrows $g: P \to A$ and $h: A \to P$ such that $gh = 1_A$. Let $\tau: E \twoheadrightarrow X$ and $\varphi: A \to X$. We want to find $\psi: A \to E$ which satisfies $\tau \psi = \varphi$. Since $P$ is projective, we can find a $\psi ' :P \to E$ such that $\tau \psi ' = \varphi g$. Finally, set $\psi := \psi ' h: A \to E$, then, 
$$
\tau \psi = \tau \psi' h = \varphi gh = \varphi.
$$
This proves the claim.

\subsection{} Show that all epis in $\mathbf{Pos}$, the category of posets, are surjections on elements, and that the one element poset is projective.\hfill \break

\iam{Raphael} Let $f:X\rightarrow Y$ be an epis of posets. Let $Z=\{\{\},\{*\}\}$ equipped with the order $\{\}\leq \{*\}$. Define the morphisms of posets $\alpha:Y\ni y\mapsto \{*\}$ and
$\beta:Y\rightarrow Z$ with $\beta(y)=\{*\}$ if and only if $y\in f(X)$. Then we have $\alpha\circ f=\beta\circ f$, and hence $\alpha=\beta$, i.e. $f(X)=Y$.\\
Let now $X=\{*\}$ be a one element poset. Let $g:E\rightarrow Y$ be an epi of posets and $f:X\rightarrow Y$ be a morphism of posets. Because an epi of posets is surjective
(by the above) we know that there is an $e\in E$ with $g(e)=f(*)$. Define $\overline{f}(*)=e$, then clearly $g(\overline{f}(*))=g(e)=f(*)$, hence $X$ is projective.
  
\subsection{} Show that sets, regarded as \emph{discrete} posets, are projective in $\mathbf{Pos}$. Give an example of a poset which is not projective. Show that every projective poset is discrete, that is a set. Conclude that $\mathbf{Set}$ is (isomorphic to) the \emph{full subcategory} of projective objects in $\mathbf{Pos}$. \hfill \break


\subsection{} Show that a homomorphisms of graphs $h:G\to H$ is monic just if it is injective on edges and vertices.\hfill \break

\iam{Marius} The claim boils down to the fact that we have two special graphs $V :=$ \fbox{$\bullet$} and $E:=$ \fbox{$\bullet \rightarrow \bullet$} that allow us to distinguish vertices and arrows in any graph. More precisely, for any graph $G$ we have isomorphisms $\alpha: \text{Hom}(V,G) \cong V(G)$ and $\beta: \text{Hom}(E,G) \cong E(G)$ in $\mathbf{Set}$, where $V(G)$ denotes the vertex set of $G$, and $E(G)$ its edge set. In fact, assigning each graph its underlying set of vertices (resp. edges) defines functors $V,E: \mathbf{Graph} \rightarrow \mathbf{Set}$. Moreover, for any $h: G \rightarrow G'$ in $\mathbf{Graph}$ the following squares commute, where $h_*(v) = h \circ v$, and $V(h)$, $E(h)$ are the maps induced on the underlying sets by $h$. (Think this through yourself, as it is not enlightening to write it down in symbols).
\begin{figure}[!h]
	\centering
\begin{tikzcd}
{\text{Hom}(V,G)} \arrow[rr, "\alpha_G"] \arrow[dd, "h_{*}"] &amp;  &amp; V(G) \arrow[dd, "V(h)"] \\
&amp;  &amp;                         \\
{\text{Hom}(V,G')} \arrow[rr, "\alpha_{G'}"]                 &amp;  &amp; V(G')                  
\end{tikzcd}
\qquad
\begin{tikzcd}
{\text{Hom}(V,G)} \arrow[rr, "\beta_G"] \arrow[dd, "h_{*}"] &amp;  &amp; E(G) \arrow[dd, "E(h)"] \\
&amp;  &amp;                         \\
{\text{Hom}(V,G')} \arrow[rr, "\beta_{G'}"]                 &amp;  &amp; E(G')                  
\end{tikzcd}
\end{figure}
This shows that the above isomorphisms are natural. In particular the functors $V$ and $E$ are representable, represented by the objects $V$ and $E$, respectively. 

We next observe that in any category $\mathcal{C}$, a morphism $h: A \rightarrow B$ is mono iff for every $C \in \mathcal{C}$ the map $h_* : \text{Hom}(C,A) \rightarrow \text{Hom}(C,B)$ is injective. (Sketch: ($\Rightarrow$) For $f \neq g \in \text{Hom}(C,A)$, $h_*(f) = h \circ f \neq h \circ g = h_*(g)$, since h monic. ($\Leftarrow$) For $f,g: T \rightarrow A$, $h: A \rightarrow B$, if $h \circ f = h_*(f)= h_*(g) = h \circ g$, then $f = g$, by injectivity). In particular, if $h: G \rightarrowtail G'$ is monic in $\mathbf{Graph}$, then the image of $h$ under both $\text{Hom}(V,-)$ and $\text{Hom}(E,-)$ is injective. By the natural isomorphisms $\alpha$ and $\beta$, this implies that $V(h)$ $(=\alpha_{G'} \circ h_* \circ \alpha_G^{-1})$, and similarly $E(h)$, are injective, as was to be shown.  

The above argument generalizes: Whenever we can represent the functor $U$ which maps object to their underlying set by some object in our category, we may conclude that monomorphisms are injective on the underlying sets. In particular, this holds for sets ($\text{Hom}(\{*\}, X) \cong X$), monoids ($\text{Hom}(\mathbb{N}, M) \cong U(M)$), groups ($\text{Hom}(\mathbb{Z}, G) \cong U(G)$), $R$-modules ($\text{Hom}(R,M) \cong U(M)$), posets ($\text{Hom}(\scalebox{0.6}{\begin{tikzcd}
\bullet \arrow[loop, distance=1.25em, in=115, out=65]
\end{tikzcd}},P) \cong U(P)$), and so on. 

\subsection{} Let $A$ be a set. Define an $A$-monoid to be a monoid $M$ equipped with a function of sets $m:A\to U(M)$ ($U(M)$ is the underlying set of the monoid $M$). A morphism of $A$-monoids 
$$h:(M,m)\to (N,n)$$
 is a monoid homomorphism which commutes the following triangle in $\mathbf{Set}$
\begin{center}
\begin{tikzcd}
U(M) \arrow[rr, "U(h)"] &amp;                                   &amp; U(N) \\
                        &amp; A \arrow[lu, "m"] \arrow[ru, "n"] &amp;     
\end{tikzcd}
\end{center} 
where $U(h)$ is the morphism $h:M\to N$ treated as the function between the underlying sets. Together with the obvious identities this forms a category $A-\mathbf{Mon}$ of $A$-monoids.

Show that the initial object in this category is the free monoid on $A$, $M(A)$. \emph{(Hint: see section 1.7 for the definition of free monoid)}\hfill \break

\subsection{} Show that in a category with binary products, the product is associative. Meaning that there is a canonical isomorphism 
$$(A\times B)\times C\cong A\times (B\times C)$$
for any triple of objects $A,B,C$.\hfill \break

\subsection{} Given a category $\mathbf{C}$ with two distinguished objects $A$ and $B$, define a category $\mathbf{C}_{A,B}$ to have objects $(X,x_1,x_2)$ with 
\begin{center}
\begin{tikzcd}
A &amp;                                        &amp; B \\
  &amp; X \arrow[lu, "x_1"] \arrow[ru, "x_2"'] &amp;  
\end{tikzcd}
\end{center}
and with arrows $f:(X,x_1,x_2)\to (Y,y_1,y_2)$ being arrows $f:X\to Y$ in $\mathbf{C}$ commuting the diagram 
\begin{center}
\begin{tikzcd}
A                                                       &amp;                                           &amp; B \\
                                                        &amp;                                           &amp;   \\
X \arrow[uu, "x_1"] \arrow[rruu, "x_2"'] \arrow[r, "f"] &amp; Y \arrow[luu, "y_1"'] \arrow[ruu, "y_2"'] &amp;  
\end{tikzcd}
\end{center}

Show that $\mathbf{C}_{A,B}$ has a terminal object if and only if $A\times B$ is well defined in $\mathbf{C}$. \hfill \break

\subsection{} In any category $\mathbf{C}$ with binary products, define the \emph{graph} of an arrow $f:A\to B$ to be the morphism $\Gamma(f)$ defined by the following universal property:
\begin{center}
\begin{tikzcd}
A &amp;                                                                                                        &amp; B \\
  &amp; A\times B \arrow[ru, "p_2"] \arrow[lu, "p_1"]                                                          &amp;   \\
  &amp; A \arrow[luu, "\mathrm{id}_A", bend left] \arrow[ruu, "f"', bend right] \arrow[u, "\Gamma(f)", dashed] &amp;  
\end{tikzcd}
\end{center}
\begin{enumerate}
\item Show that this arrow is always monic;
\item For $\mathbf{C}=\mathbf{Set}$, show that this defines a functor $\Gamma:\mathbf{Set}\to \mathbf{Rel}$. \emph{(Hint: for the definition of the category $\mathbf{Rel}$ see Exercise \ref{Rel}. To define the relation $R(f)\in A\times B$ use the image of $\Gamma(f)$)}
\end{enumerate}\hfill \break

\subsection{} \emph{(Harder)} Show that the forgetful functor $U:\mathbf{Mon}\to \mathbf{Set}$ from monoids to sets which forgets about the monoid structure is representable. Show that $U$ preserves small products.

\iam{Severin} The forgetful functor should be representable by $\operatorname{Hom}(M(*), \cdot)$, where $M(*)$ denotes the free monoid on $\{*\}$. The forgetful functor $U: \textbf{Mon} \rightarrow \textbf{Set}$ is on objects $U(M)= \vert M \vert$ (underlying set of the monoid) and $U(f) = \vert f \vert$ (underlying function on the sets). The functor $\operatorname{Hom}(M(*), \cdot)$ is given by 
\begin{align*}
\operatorname{Hom}(M(*), \cdot) : \ &amp;\textbf{Mon} \rightarrow \textbf{Set} \\
&amp;M \mapsto \operatorname{Hom}(M(*), M) \\
&amp;(f: B \rightarrow B') \mapsto \left( \operatorname{Hom}(M(*), B) \rightarrow \operatorname{Hom}(M(*), B'), g \mapsto f \circ g \right).
\end{align*}
Now we have to identify those two functors. We consider the following two maps on the level of objects.
\begin{align*}
\Psi: \operatorname{Hom}(M(*), M) \rightarrow \vert M \vert, \ \overline{f} \mapsto \vert \overline{f} \vert (i(*))
\end{align*}
where $i: \{*\} \rightarrow M(\{*\})$ as in the UMP of the free monoid (see Awodey p. 17). And 
\begin{align*}
\Phi: \vert M \vert \rightarrow \operatorname{Hom}(M(*), M), m \mapsto \overline{f}_m
\end{align*}
where $\overline{f}_m$ is the unique monoid morphism we get from the UMP of the free monoid with respect to the function $f_m : \{*\} \rightarrow \vert M \vert, \ f_m(*) = m$. We have that $\Psi, \Phi$ are mutually inverse and thus $\vert M \vert$ is isomorphic to $\operatorname{Hom}(M(*), M)$ in the category of sets. Indeed, we have
\begin{align*}
(\Psi \circ \Phi) (m) = \vert \overline{f}_m \vert (i(*)) = f_m(*) = m.
\end{align*}
On the other hand, we have
\begin{align*}
(\Phi \circ \Psi)(\overline{f}) = \overline{f}_{\vert \overline{f} \vert(i(*))} = \overline{f},
\end{align*}
where we used in the last line the UMP for the free monoid ($\vert \overline{f}_{\vert \overline{f}\vert(i(*))} \vert(i(*))= \vert \overline{f} \vert(i(*))$ and thus, by the uniqueness part of the UMP $\overline{f}_{\vert \overline{f} \vert(i(*))} = \overline{f}$).
Feel free to flesh out this hand-wavy sketch.
\end{document}
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