Let $f:X\longrightarrow Y$ be a morphism of topological spaces. I want to prove, that the inverse-image functor of (set-valued) presheaves
\begin{equation}
f^{-1} : \mathrm{PSh}(Y)\longrightarrow\mathrm{Psh}(X), \mathcal{F}\mapsto f^{-1}\mathcal{F}
\end{equation}
is a functor. Let $\mathrm{Op}(X)$ be the index-category of open sets of $X$. Then for open $U_2\subset U_1\subset X$ we get full subcategories of $\mathrm{Op}(Y)$, namely
\begin{equation}
J_1 =\{V\in\mathrm{Op}(Y) : V\supset f(U_1)\}, \quad J_2 = \{V\in\mathrm{Op}(Y) : V\supset f(U_2)\}
\end{equation}
with $J_1\subset J_2$, because $f(U_2)\subset f(U_1)$. Now for a presheaf $\mathcal{F}:\mathrm{Op}(X)\longrightarrow\mathrm{Set}$ the restriction map
\begin{equation}
f^{-1}{\mathcal{F}}(U_1) = \underset{V\supset f(U_1)}{\mathrm{colim}}\mathcal{F}(V) = \mathrm{colim}_{J_1}(\mathcal{F})\longrightarrow\mathrm{colim}_{J_2}(\mathcal{F}) = \underset{V\supset f(U_2)}{\mathrm{colim}}\mathcal{F}(V) = f^{-1}\mathcal{F}(U_2)
\end{equation}
projects an equivalence class $[A, a]_{J_1}$ of $a\in\mathcal{F}(A)\subset\coprod_{V\in J_1}\mathcal{F}(V)$ to an equivalence class $[A,a]_{J_2}$ of the same element. Further for a morphism $\alpha : \mathcal{F}\longrightarrow\mathcal{G}$ of presheaves, the map
\begin{equation}
f^{-1}\mathcal{F}(U_1) = \mathrm{colim}_{J_1}(\mathcal{F})\longrightarrow\mathrm{colim}_{J_1}(\mathcal{G}) = f^{-1}\mathcal{G}(U_1)
\end{equation}
sends an equivalence class $[A,a]_{J_1}$ to $[A,\alpha(A)(a)]_{J_1}$. To see, that $f^{-1}(\alpha) : f^{-1}\mathcal{F}\longrightarrow f^{-1}\mathcal{G}$ is a morphism of presheaves, we have to check, that the diagram
$\require{AMScd}$
\begin{CD}
f^{-1}\mathcal{F}(U_1) = \mathrm{colim}_{J_1}(\mathcal{F}) @>{\mathrm{colim}(\alpha)}>> \mathrm{colim}_{J_1}(\mathcal{G}) = f^{-1}\mathcal{G}(U_1)\\
@VVV @VVV\\
f^{-1}\mathcal{F}(U_2) = \mathrm{colim}_{J_2}(\mathcal{F}) @>{\mathrm{colim}(\alpha)}>> \mathrm{colim}_{J_2}(\mathcal{G}) = f^{-1}\mathcal{G}(U_2)
\end{CD}
commutes. But by construction the two ways
\begin{equation}
[A,a]_{J_1}\mapsto [A,\alpha(A)(a)]_{J_1}\mapsto [A,\alpha(A)(a)]_{J_2}, \quad [A,a]_{J_1}\mapsto [A,a]_{J_2}\mapsto [A,\alpha(A)(a)]_{J_2}
\end{equation}
are trivially the same.
Now I have two questions:
- Is it possible to proof the above just with universal properties, without using explicit construction of colimits in $\mathrm{Set}$?
- (maybe another formulation of 1)) Is there a generalization, that for arbitrary-valued functors $\mathcal{F} : I\longrightarrow C$ and subcategories $J_1\subset J_2\subset I$ the canonical diagram
\begin{CD}
\mathrm{colim}_{J_1}(\mathcal{F}) @>{\mathrm{colim}(\alpha)}>> \mathrm{colim}_{J_1}(\mathcal{G})\\
@VVV @VVV\\
\mathrm{colim}_{J_2}(\mathcal{F}) @>{\mathrm{colim}(\alpha)}>> \mathrm{colim}_{J_2}(\mathcal{G})
\end{CD}
commutes?
Best Answer
The answer to 2) (and so 1) is yes. To see it, let's recall what the maps are.
Take a colimit cocone of $\mathcal{F}$ over $J_2$. Its restriction to $J_1$ is a cocone of $\mathcal{F}$ again. So it induces a unique morphism from $\mathrm{colim}_{J_1}(\mathcal{F}) \to \mathrm{colim}_{J_2}(\mathcal{F})$, this is a construction-free definition of the vertical map of your diagram.
The horizontal maps are constructed in the following way: given a natural transform $\alpha: \mathcal{F} \to \mathcal{G}$, take the colimit cocone of $\mathcal{G}$ and precompose it by the natural transform $\alpha$. By naturality of $\alpha$ this is a cocone under $\mathcal{F}$, and so it again induces a unique map from $\mathrm{colim}(\mathcal{F}) \to \mathrm{colim}(\mathcal{G})$.
Now compare the constructions of the two composites maps in your diagram. Let's look first at the one going down then left: i.e $\mathrm{colim}(\alpha) \circ \rho_F$ where $\rho_F$ is the restriction. To build the composite. First take the colimit cocone of $\mathcal{G}$ over $J_2$, precompose with $\alpha$, induce colimit, then restrict, then induce colimit again. By the universal property of the colimit, this map is in fact the unique map $\mathrm{colim}_{J_1}(\mathcal{F}) \to \mathrm{colim}_{J_2}(\mathcal{G})$ that makes the cocone obtained from the colimit cocone of $\mathcal{G}$ by precomposition by $\alpha$ and then restriction to $J_1$.
But precomposition by $\alpha$ and restriction to $J_1$ commute (in the sense that first restricting the diagram then precomposing with a restricted alpha is the same as precompose by $\alpha$ then restrict the whole thing.) So this map is also the map you get from the other composition, which you can check satisfies the same property because precompostion and restriction commute.