Inverse image of sheaves

Question

I am trying to prove the following: $$f^{-1}(g^{-1}(\mathscr{H})) = (g \circ f)^{-1}(\mathscr{H})$$ where $\mathscr{H}$ is a presheaf on some topological space $Z$ and $f: X \rightarrow Y$, $g : Y \rightarrow Z$ are continuous maps between topological spaces. My definition of inverse image of a sheaf is the following: given $\mathscr{F}$ a sheaf on $Y$ and $f : X \rightarrow Y$ a continuous map between topological spaces we define $$f^+\mathscr{F}(U) = \lim_{ \begin{array}{c} \longrightarrow\\ V \supseteq f(U)\\ V \subseteq Y \text{ open} \end{array}} \mathscr{F}(V) $$ and we define $f^{-1}\mathscr{F}$ to be the sheafification of $f^+\mathscr{F}$. I understand that $f^+(g^+(\mathscr{F})) = (g \circ f)^+(\mathscr{F})$, hence: $(g \circ f)^{-1}(\mathscr{F}) = \mathscr{G}$, where with $\mathscr{G}$ I mean the sheafification of $f^+(g^+(\mathscr{F})$. Now by the universal property of sheafification we obtain that $\mathscr{G} = f^{-1}(g^+(\mathscr{F}))$ and we have a natural morphism $f^{-1}(g^+(\mathscr{F})) \rightarrow f^{-1}(g^{-1}(\mathscr{H}))$ which is the map induced by $f^+(i)$, where $i: g^+(\mathscr{F}) \rightarrow g^{-1}(\mathscr{F})$ is the sheafification map. I don't understand why that map must be an isomorphism (I can't use the characterization of the stalks of the presheaf $f^+(-)$ or of $f^{-1}(-)$) and I don't find another way to show the isomorphism at the beginning. Thank you.

Answer 1

I find proving equalities like this way more "intuitive" using the universal properties. You might want to take the following approach.

Given $f:X\to Y$ and a sheaf $\mathscr{F}$ on $X$, the \emph{direct image} sheaf $f_*\mathscr{F}$ (on $Y$) is defined as $f_*\mathscr{F}(U)=\mathscr{F}(f^{-1}U)$. In the language of category theory the functor $f^{-1}$ (that you defined) is the left-adjoint to direct image functor $f_*$. More explicity, given any sheaf $\mathscr{F}$ on $X$ and any sheaf $\mathscr{G}$ on $Y$ one has a natural bijection of sets: $$ \alpha: \mathrm{Hom}_Y(\mathscr{G}, f_*\mathscr{F})\to \mathrm{Hom}_{X}(f^{-1}\mathscr{G}, \mathscr{F}) $$ where by $\mathrm{Hom}_X(\mathscr{F_1}, \mathscr{F}_2)$ I mean the set of morphisms of sheaves from $\mathscr{F}_1$ to $\mathscr{F}_2$. You might want to first find such $\alpha$. This is called the adjoint property (I'll provide hints on how to do this at the end.

Using the adjoint property prove that thre is a natural bijection $$ \beta:\mathrm{Hom}_X (f^{-1}(g^{-1}\mathscr{H}), \mathscr{F)}\to \mathrm{Hom}_X ((f\circ g)^{-1}\mathscr{H}, \mathscr{F}) $$ for any sheaf $\mathscr{H}$ on $Z$ and any sheaf $\mathscr{F}$ on $X$. In particular let $\mathscr{F}=f^{-1}(g^{-1}\mathscr{H})$, with its bijeciton

$$ \beta:\mathrm{Hom}_X (f^{-1}(g^{-1}\mathscr{H}), f^{-1}(g^{-1}\mathscr{H}))\to \mathrm{Hom}_X ((f\circ g)^{-1}\mathscr{H}, f^{-1}(g^{-1}\mathscr{H})) $$ Now choose the identity morephism $\mathrm{id}\in \mathrm{Hom}_X (f^{-1}(g^{-1}\mathscr{H}), f^{-1}(g^{-1}\mathscr{H}))$. What is $\beta(\mathrm{id})$ now? What does say about $(f\circ g)^{-1}$ and $f^{-1}\circ g^{-1}$?

Adjoint property: Consider a morphism $\phi: \mathscr{G}\to f_*\mathscr{F}$, meaning for each open set $V\subset Y$, one has $\phi(V): \mathscr{G}(V)\to \mathscr{F}(f^{-1}(V))$ compatible with restriction. Take an open set $U\subset X$, and suppose $V\subset Y$ is such that $f(U)\subset V$, then $U\subset f^{-1}(V)$, therefore using a restriction map we have a function $\mathscr{F}(f^{-1}(V))\to \mathscr{F}(U)$. Gathering all of this you will find a family of maps $$ \psi_V: \mathscr{G}(V)\to \mathscr{F}(U) $$ for all $V\supset f(U)$. By universal property of direct limit then there exists a map, compatible with all these maps, $$ \omega(U): f^+\mathscr{G}(U)\to \mathscr{F}(U) $$ I leave it to you to check this naturally extends to a morphism $\omega: f^{-1}\mathscr{G}\to \mathscr{F}$ (use universal propert of sheafification). Define $\alpha(\phi)=\omega$.

Going in the reverse direction, take a morphism $\omega(U): f^{-1}\mathscr{G}(U)\to \mathscr{F}(U)$. Take an open set $V\subset Y$ and let $U=f^{-1}(V)$. Then $f^{-1}\mathscr{G}(U)=\mathscr{G}(V)$, and using $\omega$, we a map $\phi(V): \mathscr{G}(V)\to f_*\mathscr{F}(V)$. Show that $\phi$ is indeed a morphism of sheaves. Then prove that this map is inverse to $\alpha$.