# Morphisms of sheaves

My question concerns direct/inverse image of sheaves and their properties. Let $\mathfrak{R}$, $\mathfrak{S}$, $\mathfrak{T}$ three sheaves of groups, over topological spaces $X$, $Y$ and $Z$ respectively, and suppose that there exist two continuous functions $f:X\longrightarrow Y$ and $g:Y\longrightarrow Z$. My goal is to naturally define a morphism of sheaves $\mathfrak{R}\longrightarrow (g\circ f)^{-1} (\mathfrak{T})$, given that two morphisms $\mathfrak{R}\longrightarrow f^{-1}\mathfrak{S}$ and $\mathfrak{S}\longrightarrow g^{-1}\mathfrak{T}$ exist. Is there a natural way of defining such a morphism? My biggest problem with this definition is that I am not sure weather $(f_{*}f^{-1})(\mathfrak{S})=\mathfrak{S}$.

• There is always a map $\mathcal G \to f_*f^{-1}\mathcal G$, but usually this isn't an isomorphism. Take $X$ to be a point, for example. – Hoot Jan 13 '15 at 15:49
• That's what I taught. Is there a theorem that gives sufficient conditions for $\mathcal{G}\rightarrow f_{*}f^{-1}\mathcal{G}$ to be an isomorphism? – Fq00 Jan 14 '15 at 10:16

Given $\mathfrak{R}\longrightarrow f^{-1}\mathfrak{S}$ and $\mathfrak{S}\longrightarrow g^{-1}\mathfrak{T}$ gives you $\mathfrak{R}\longrightarrow f^{-1} \mathfrak{S}\longrightarrow f^{-1} (g^{-1}\mathfrak{T}) = (g\circ f)^{-1}\mathfrak{T}$, which is the wanted arrow. Of course, the arrow $f^{-1} \mathfrak{S}\longrightarrow f^{-1} (g^{-1}\mathfrak{T})$ is obtained from the arrow $\mathfrak{S}\longrightarrow g^{-1}\mathfrak{T}$ by functoriality of $f^{-1}$.
Let me detail a bit the inverse image sorite. This is general, say you have $f : X \rightarrow Y$ continuous. How is defined the functor $f^{-1}$ from the category of sheaves of abelian groups of $Y$ to the category of sheaves of abelian groups on $X$ ? First, how does it act on objects ? Let $\mathscr{F}$ be a sheaf of abelian groups on $Y$. Consider the following presheaf of abelian groups $I_{\mathscr{F}} : U\mapsto {\varinjlim}_{f(U)\subseteq V}\mathscr{F}(U)$. Then the sheaf $f^{-1} \mathscr{F}$ is simply the sheaf associated to this presheaf by the sheafification functor. How does our inverse image functor act on arrows ? For this, let $u : \mathscr{F} \rightarrow \mathscr{G}$ a morphism of sheaves of abelian groups on $Y$. Interpreting presheaves as contravariant functors from the category of open sets of $Y$ with inclusions as arrows and interpreting sheaves in the same way plus the equalizer conditions, $u$ is simply a morphism of functor, as so-called natural transformation. Or also a morphism of directed systems of abelian groups. This allows to define a morphism $I_{u} : I_{\mathscr{F}} \rightarrow I_{\mathscr{G}}$. Applying to it the sheafification functor gives you $f^{-1} u$, and finished the definition of the inverse image functor. Valid of course for the category of sheaves of abelian groups. For $\mathscr{O}_Y$-modules the definition is not the same : the inverse image is noted $f^{\star}$ and is defined as $f^{\star} \mathscr{M} = (f^{-1} \mathscr{M}) \otimes_{f^{-1} \mathscr{O}_Y} \mathscr{O}_X$ for any $\mathscr{O}_Y$-Module $\mathscr{M}$.
• Thank you for your reply. Can you please give me the explicit construction of the map $(g\circ f)^{-1} \mathfrak{T}$ on the open subsets? – Fq00 Jan 14 '15 at 10:12
• Edited my answer to address the question of your last comment. Note that the key is simply to know how how the functor $f^{-1}$ is defined, as after you apply it to an arrow, that you then compose with another arrow to get the final arrow. – Olórin Jan 14 '15 at 15:23