Exercises.tex

\subsection*{Pull-back Functor} In the remainder of this week's exercises we can look more deeply at the concept of slice category and pull-back functor. In Awodey the definition of slice category is in Section 1.6, while some details of the pull-back functor are in Section 5.3.

\begin{defn}(Slice Category) Let $\mathcal{C}$ be a category and $A$ a distinguished object. Define the slice category $\mathcal{C}\downarrow A$ as the category whose objects are morphisms with codomain $A$, such as $f$ or $g$ below, and morphisms are commutative diagrams such as 
\begin{center}
\begin{tikzcd}
X \arrow[rr, "g"] \arrow[rd, "\alpha"] &                   & A \\
                                       & Y \arrow[ru, "f"] &  
\end{tikzcd}
\end{center} 
which we typically just label by $\alpha:f\to g$.
\end{defn}

\begin{ex} Let $\mathbf{Emb}_d$ be the category of smooth $d$ dimensional manifolds and open embeddings between them. Choose a manifold $M$, then $\mathbf{Emb}_d\downarrow M$ is the poset of open subspaces of $M$, sometimes called $\mathrm{Open}(M)$.
\end{ex}

Now consider a diagram of this type:
\begin{center}
\begin{tikzcd}
B \arrow[rr, "h"] &  & A                 \\
                  &  &                   \\
                  &  & Y \arrow[uu, "f"]
\end{tikzcd}
\end{center}
Both $f$ and $h$ can be viewed as objects of $\mathcal{C}\downarrow A$, but our immediate objective is to fix $B$ and ``transfer'' $f$ to an object of the slice category $\mathcal{C}\downarrow B$. This can be done if $\mathcal{C}$ has pull-backs using the following diagram 
\begin{center}
\begin{tikzcd}
B \arrow[rr, "h"]                                                             &    & A                 \\
                                                                              & \; &                   \\
B\times_A Y \arrow[rr] \arrow[uu, "h^*f"] \arrow[ru, "\urcorner" description, phantom, near start] &    & Y \arrow[uu, "f"]
\end{tikzcd}
\end{center}
where $h^*f$ is called the pull-back of $f$ by $h$.

\begin{ex} Returning to our previous example, $\mathbf{Emb}_d$ has all pull-backs, so given a continuous map $h:M\to N$ one obtains a monotone map (of posets) 
\begin{align*}
h^*:\mathrm{Open}(N)&\to \mathrm{Open}(M)\\
U&\longmapsto h^{-1}(U)
\end{align*}
which acts by ``pulling back'' open subspaces from $N$ to $M$.
\end{ex}

\subsection{} Let $\mathcal{C}$ be a category with pull-backs and $h:B\to A$ a morphism. Prove that this defines a functor $$h^*: (\mathcal{C}\downarrow A)\to (\mathcal{C}\downarrow B)$$ using the description above. (\emph{Hint:} use the two pull-back lemma.)\hfill \break

\subsection{} Let $\mathcal{C}$ be a small category with pull-backs and $\mathbf{Cat}$ be the (1-)category of small categories. Show, giving details, that there exists a functor 
\begin{align*}
\mathcal{C}&\to \mathbf{Cat}\\
A&\mapsto \mathcal{C}\downarrow A\\
(h:B\to A)&\mapsto (h^*: (\mathcal{C}\downarrow A)\to (\mathcal{C}\downarrow B))
\end{align*}
Speculate on the importance of this functor.\hfill \break

\section{Week 5}

\subsection{} Show that, for any three objects $A,B,C$ in a cartesian closed category, there are isomorphisms:
\begin{enumerate}
\item $(A\times B)^C\cong A^C\times B^C$\\
\iam{Arno} We want to show, that $A^C\times B^C$ satisfies the exponential condition, which implies the isomorphism. \\
First we want to define an evaluation map $ev: A^C\times B^C \times C \to A\times B$. To do so, we have to have to give a map into $A$ and $B$ respectively. \\
We allready have evaluation maps $ev_{A}: A^C\times C \to A$ and $ev_{B}: B^C\times C \to B$. We take $ev = <ev_{A} \circ \pi _{A^C\times C},ev_{B}\circ \pi _{B^C \times C}>$.\\
Because $X \times Y \times Z \cong (X \times Y) \times Z \cong X \times (Y \times Z)$ we have the maps $\pi _{A^C \times C}$ and $\pi _{B^C \times C}$, so the defined map makes sense.\\
Now $\forall g: X \times C \to A \times B$ $g$ splits into $g_{A}: X\times C \to A$ and$ g_{B}:X\times C \to B$ from which we get from the exponential property unique maps $\tilde{g_{A}}:X\to A^C$ and $\tilde{g_{B}}:X\to B^C$ such that $g_{i}=ev_{i}\circ \tilde{g_{i}}\times id_{C}$\\
Thus we see that there is a unique morphism $\tilde{g}=\tilde{g_{A}}\times \tilde{g_{B}}$ such that $g=ev \circ (\tilde{g}\times id_{C})$\\
\item $(A^B)^C\cong A^{(B\times C)}$\\
\iam{Andreas} Since it's CCC there must exist and objects $(A^B)^C,A^{B\times C}$ and $\operatorname{ev}_{A^B}:(A^B)^C\times C \to A^B$ and $\tilde{q}:A\to A^{B\times C}$and $\operatorname{ev}_{A,2}:A^{(B\times C)}\times (B\times C) \to A$, and $\tilde{g}_{A^B}:X\to (A^B)^C$ and $\tilde{g}_{A,2}:X\to A^B$\\
So $g:(A^B)^C\to A^{(B\times C)}$ with
$g(\cdot):=\tilde{q}\circ\operatorname{ev}_B(\operatorname{ev}_{A^B}(\cdot,1_C),1_B)$ and $g^{-1}:A^{(B\times C)}\to (A^B)^C$ with
$g^{-1}(\cdot):=\tilde{g}_{A^B}\circ\operatorname{ev}_{A,2}(\cdot,1_{B\times C})$\\
\end{enumerate}

\subsection{} Is the category of monoids cartesian closed?\hfill \break

\subsection{} How is the category of abelian groups cartesian closed? Describe its exponential functor.\hfill \break

\subsection{} Show that, for any objects $A,B$ in a cartesian closed category there is a bijective correspondence
\begin{align*}
\Hom(1,B^A)\cong \Hom(A,B)
\end{align*}

\subsection{} Prove that, for any cartesian closed category $\mathcal{C}$ with a fixed base object $A$, the operation 
\begin{align*}
A^{(-)}:\mathcal{C}^{op}\to \mathcal{C}
\end{align*}
describes a functor.\hfill\break

\subsection{} Let $\mathcal{C}$ be a cartesian closed category with coproducts, show that it is distributive in the sense that there exists a (natural) isomorphism
\begin{align*}
(A\times C)+(B\times C)\cong (A+B)\times C
\end{align*}


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