TITLE: About the presheaf used to define the inverse image sheaf.
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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.