Let $X$ and $Y$ be topological spaces, and $\varphi:X\to Y$ a continuous map. Let $\mathbf{X}$ and $\mathbf{Y}$ be the categories corresponding to the open sets of $X$ and $Y$ with arrows given by reverse inclusion. Then a presheaf on $X$ is a functor $\mathcal{F}:\mathbf{X}\to\mathbf{Set}$, and the category $\mathbf{PShv}(\mathbf{X})$ of presheaves on $X$ is just the functor category $\mathbf{Set}^\mathbf{X}$.
We obtain a functor $\Phi:\mathbf{Y}\to\mathbf{X}$ by setting $\Phi(U)=\varphi^{-1}(U)$. Then the direct image functor $I:\mathbf{PShv}(\mathbf{X})\to\mathbf{PShv}(\mathbf{Y})$ is given by setting $I(\mathcal{F})=\mathcal{F}\circ\Phi$.
However I can't seem to give such a clean construction of the inverse image functor $J:\mathbf{PShv}(\mathbf{Y})\to\mathbf{PShv}(\mathbf{X})$. I know that for a presheaf $\mathcal{G}$ on $Y$ and an open $U\subseteq X$ we set $$\varphi^{-1}\mathcal{G}(U)=\mathop{\lim_{\longrightarrow}}_{V\supseteq\varphi(U)}\mathcal{G}(V)$$
I was hoping to find a more "categorical" description though, since, at least to me, the functoriality of $J$ is not obvious from this in the same way that it is with $I$. I know that $J$ is the left adjoint of $I$, so I feel like there should be another construction which makes this clearer.
Wikipedia says that the functoriality follows from the universal property of colimits. I think the colimit is taken over the images under $\mathcal{G}$ of the $V\subseteq Y$ such that there is an arrow $\Phi(V)\to U$ in $\mathbf{X}$, but I'm struggling to see how this shows $J$ is a functor.
Any help would be much appreciated.
Best Answer
Ah, I just realized: $\mathcal F \overset{\mathrm{def}}{=} \mathcal G$.
You basically want to construct a morphism $\Gamma(V, \varphi^{-1} \mathcal F) = \varinjlim_\limits{U \supseteq \varphi(V)} \Gamma(U,\mathcal F) \to \varinjlim_\limits{U' \supseteq \varphi(V')}\Gamma(U', \mathcal F) = \Gamma(V', \varphi^{-1}\mathcal F)$ for any inclusion $V' \subseteq V$ of open subsets.
Now the universal property of the colimit used to compute $\Gamma(V,\varphi^{-1}\mathcal F)$ tells you how to obtain morphism starting at $\Gamma(V,\varphi^{-1}\mathcal F)$.
Namely: You have to give a morphism $\Gamma(U,\mathcal F) \to \varinjlim_\limits{U' \supseteq \varphi(V')}\Gamma(U', \mathcal F) = \Gamma(V', \varphi^{-1}\mathcal F)$ for any open $U \supseteq \varphi(V)$ compatible with the transition maps (which are of course given via restriction).
Now just note that since $V' \subseteq V$ one certainly has $U \supseteq \varphi(V) \supseteq \varphi(V')$ so that $\Gamma(U,\mathcal F)$ is a part of the second colimit, i.e. comes with a canonical map into $\Gamma(U,\mathcal F) \to \Gamma(V',\varphi^{-1}\mathcal F)$ as desired.
Then you have to check that these maps are compatible with restrictions so that they induce the desired morphism $\Gamma(V,\varphi^{-1}\mathcal F) \to \Gamma(V', \varphi^{-1} \mathcal F)$.
$\textit{Uniqueness}$ of the morphism induced on the colimit then tells you that this construction behaves well with respect to inclusions $V'' \subseteq V' \subseteq V$, i.e. is functorial.