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.

  • $\begingroup$ Why can't you use the characterization of the stalks ? Is this part of an exercise ? If you are looking for another proof, can you use Yoneda lemma and its corollaries ? For example $f^{-1}g^{-1}$ and $(g\circ f)^{-1}$ are both left adjoint to $g_*f_*=(g\circ f)_*$ hence are isomorphic. $\endgroup$
    – Roland
    Sep 21, 2017 at 20:53
  • $\begingroup$ I can't use the characterization of the stalks because the book on which I'm studying obtains the characterization by this property and I'd like to understand it's reasoning. Also the adjointness is done later but I didn't know about the proof by Yoneda's Lemma, it's very nice! Thank you! $\endgroup$
    – Federico
    Sep 23, 2017 at 9:25
  • 1
    $\begingroup$ Well, this works also for presheaves (for you other post), and you can also prove the following result (with $a$ being the sheafification functor) $af^+=f^{-1}a$. This implies your result : for a sheaf $\mathcal{F}$, $$(gf)^{-1}\mathcal{F}=a(gf)^+\mathcal{F}=af^+g^+\mathcal{F}=f^{-1}ag^+\mathcal{F}=f^{-1}g^{-1}a\mathcal{F}=f^{-1}g^{-1}\mathcal{F}$$ and this is natural in $\mathcal{F}$. $\endgroup$
    – Roland
    Sep 23, 2017 at 10:04
  • $\begingroup$ Perfect, I'll try to prove that equality! $\endgroup$
    – Federico
    Sep 23, 2017 at 10:07
  • $\begingroup$ Well, without Yoneda, it will be difficult. But this is immediate with Yoneda : both are left adjoint to $f_*U=Uf_*$ (with $U$ the forgetful functor from sheaves to presheaves), so they are isomorphic. $\endgroup$
    – Roland
    Sep 23, 2017 at 10:11

1 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 $\alpha$.

  • $\begingroup$ Very nice proof, I knew about the adjointness but I didn't know about this proof (which it is a wise use of Yoneda's Lemma). Thank you very much! The problem as I said in the comment above is that the book on which I'm studying proves the adjointness later so I think there is another way to obtain the isomorphism from the equality $f^+(g^+(\mathscr{H})) = (g \circ f)^+(\mathscr{H})$. $\endgroup$
    – Federico
    Sep 23, 2017 at 9:27

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