About the presheaf used to define the inverse image sheaf.

Question

Let $f \colon X \to Y$ be continuous and $\mathcal{F}$ be a sheaf on $Y$. Then the inverse image sheaf $f^*\mathcal{F}$ is defined to be the sheafification of the presheaf on $X$ given by $$ U \mapsto \text{colim}_{f(U) \subseteq V} \;\mathcal{F}(V). $$

My question: How do we construct the restriction maps of this presheaf (preferable in a category theoretical way) and prove that this actually defines a presheaf?

I know that the colimit is functorial in the sense that in general for functors $F,G \colon \mathcal{I} \to \mathcal{C}$ and a natural transformation $\Phi \colon F \to G$ there is a morphism $\text{colim}_{i \in I}\,{\Phi}$ making the diagrams $$ \text{colim}_{\mathcal{i \in I}}\; F(i) \xrightarrow{\text{colim}_{i \in I}\,{\Phi}} \text{colim}_{\mathcal{i \in I}}\; G(i)$$ $$ \uparrow \hspace{4cm} \uparrow$$ $$ F(i_0) \hspace{.75cm}\xrightarrow{\hspace{.35cm}\Phi_{i_0}\hspace{.35cm}}\hspace{.75cm} G(i_0)$$ commute for all $i_0 \in \mathcal{I}$. Is there something similar for "index set changes" that i can use here?

Answer 1

$\DeclareMathOperator{\colimit}{colim}$ If $U'\subset U$, then for any $V$ containing $f(U)$, $f(U')$ must also be contained in $V$.

Now there are natural maps $\mathcal F(W)\to\colimit_{f(U')\subset V'}\mathcal F(V')$ for any $W\supset f(U')$. In particular, we have maps $$\mathcal F(V)\to \colimit_{f(U')\subset V'}\mathcal F(V')$$ for each $V\supset f(U)$. Moreover, these maps are compatible with the restriction maps between the $\mathcal F(V)$.

Hence, by the universal property of colimit, there is a unique map $$\colimit_{f(U)\subset V}\mathcal F(V)\to\colimit_{f(U')\subset V'}\mathcal F(V')$$ making the appropriate diagram commute. This is the desired restriction map. It is a slightly tedious verification using the naturality properties of the map we have constructed to verify that this indeed makes our map into a presheaf.

Answer 2

If $U \subseteq U'$ then to get a map $$\mathrm{colim}_{f(U') \subseteq V'} \ \mathcal F(V') \to \mathrm{colim}_{f(U) \subseteq V} \ \mathcal F(V)$$ we use the universal property of $\mathrm{colim}_{f(U') \subseteq V'} \ \mathcal F(V')$. So we just need a coherent system of maps $$\mathcal F(V') \to \mathrm{colim}_{f(U) \subseteq V} \ \mathcal F(V)$$ for all $V'$ containing $f(U')$. But if $V'$ contains $f(U')$ then it contains $f(U)$ and so $V'$ is in the index set over which $\mathrm{colim}_{f(U) \subseteq V} \ \mathcal F(V)$ is defined. As such there is a natural map $\mathcal F(V') \to \mathrm{colim}_{f(U) \subseteq V} \ \mathcal F(V)$ and you can easily show that these maps, defined for all $V'$, satisfy the necessary commutativity relations.