Characterisation isomorphisms of ringed spaces: $(f,f^\flat)$ iso iff $(f,f^\sharp)$ iso?

Question

So I was trying to understand isomorphisms of ringed spaces, looking for a characterization of them. I'll explain what I've found out already and what I don't know yet. Before, I will set some notations and state the pushforward-pullback adjunction of sheaves along a continuous function.

If $X$ is a space and $\mathsf{C}$ is some category, we denote by $\mathsf{Sh}_\mathsf{C}(X)$ the category of $\mathsf{C}$-sheaves over $X$. In the following, we suppose $\mathsf{C}$ is cocomplete (i.e., it admits all colimits. But we may just assume $\mathsf{C}=\mathsf{Set}$ and consider only sheaves of sets, if that makes us more comfortable).

Let $Y$ be another space. If $f:X\to Y$ is a continuous map, there are functors $$f_*:\mathsf{Sh}_\mathsf{C}X\rightleftarrows\mathsf{Sh}_\mathsf{C}Y:f^*,$$ where $f_*$ (resp., $f^*$) is called the pushforward (resp., the pullback) along $f$ (see for example the section on continuous maps and sheaves of Stacks Project for reference. There on SP they actually denote $f^*$ as $f^{-1}$. In order for $f^*$ to exist, we need cocompleteness for $\mathsf{C}$).

There is an adjunction $$ \mathsf{Sh}_{\mathsf{C}}(X)(f^{*}\mathcal{G},\mathcal{F})\cong\mathsf{Sh}_{\mathsf{C}}(Y)(\mathcal{G}, f_*\mathcal{F}),\tag{1}\label{eq:adj} $$ where $\mathcal{F}\in\mathsf{Sh}_\mathsf{C}X$ and $\mathcal{G}\in\mathsf{Sh}_\mathsf{C}Y$ (for a reference, see the link above, specially Lemma 6.21.6 and comments between lemmas 6.21.4 and 6.21.5).

Characterization of isomorphisms of ringed spaces: Recall that a morphism of ringed spaces $(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ can be specified in two ways: either in the form $(f,\psi^{\flat})$ or $(f,\psi^\sharp)$, where $\psi^\flat$ and $\psi^\sharp$ are adjunct. We say that $(f,\psi^{\flat})$ is an isomorphism if there exists $(g,\varphi^{\flat}):(Y,\mathcal{O}_Y)\to(X,\mathcal{O}_X)$ such that $(g,\varphi^\flat)\circ(f,\psi^\flat)=1_{(X,\mathcal{O}_X)}$ and $(f,\psi^\flat)\circ(g,\varphi^\flat)=1_{(Y,\mathcal{O}_Y)}$ (see this question to understand how to compose morphisms of ringed spaces). In a similar fashion, it is possible to define what it means for $(f,\psi^\sharp)$ to be an isomorphism. I am interested in showing the proving result:

Proposition. Let $(f,\psi^\flat)$ or $(f,\psi^\sharp):(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ be a morphism of ringed spaces. The following are equivalent:

1. $(f,\psi^\flat)$ is an isomorphism.
2. $f$ is a homeomorphism and $\psi^\flat$ is an isomorphism of sheaves.
3. $(f,\psi^\sharp)$ is an isomorphism.
4. $f$ is a homeomorphism and $\psi^\sharp$ is an isomorphism of sheaves.

Furthermore, in the case these equivalent statements hold, we have $$ (f,\psi^\flat)^{-1}=(f^{-1},(f^{-1})_*[(\psi^\flat)^{-1}])\\ (f,\psi^\sharp)^{-1}=(f^{-1},(f^{-1})^*[(\psi^\sharp)^{-1}]), $$ where $g_*$ denote the pushforward and $g^*$ the pullback of sheaves along the continuous function $g$ (which is a functor between some sheaves categories).

The implication $(1\Rightarrow 2)$ should be clear. For a proof of $(2\Rightarrow 1)$, see for example this answer. Equivalence $(3\Leftrightarrow 4)$ can also be proven by an analogous argument (for $(4\Rightarrow 3)$, one must also note that, from the definition of the pullback of sheaves, if $f$ is a homeomorphism then we have $f^*\mathcal{O}(U)=\mathcal{O}_Y(f(U))$, for $U\subset X$ open). On the other hand, the proofs of the formulas of the inverses of $(f,\psi^\flat)$ and $(f,\psi^\sharp)$ are an immediate computation using the fact that an inverse is unique, and the facts that $h_*$ and $h^*$ are functors and that the assignments $h\mapsto h_*$ and $h\mapsto h^*$ are functorial.

But I was having trouble showing ([$1$ or $2$] $\Leftrightarrow$ [$3$ or $4$]). Does anyone know how to prove this?

Answer 1

If $f$ is a homeomorphism, the adjunction $(1)$ can be made explicit.

Let $\varphi^\flat:\mathcal{G}\to f_*\mathcal{F}\in\mathsf{Sh}_\mathsf{C}(Y)$ be a morphism of sheaves over $Y$, where $\mathcal{F}\in\mathsf{Sh}_\mathsf{C}X$ and $\mathcal{G}\in\mathsf{Sh}_\mathsf{C}Y$. Call $\varphi^\sharp:f^*\mathcal{G}\to\mathcal{F}$ to the adjoint of $\varphi^\flat$ under the adjunction $(1)$. Then, if $f$ is a homeomorphism, these two are explicitly related by $$\tag{2}\label{eq} \varphi^\flat=f_*\varphi^\sharp. $$ A proof of \eqref{eq} can be given by

• relating $\varphi^\flat$ with $\varphi^\sharp$ using the proof of Lemma 6.21.8, where they make the adjunction $(1)$ explicit.
• using that $f_*f^*=1_{\mathsf{Sh}_\mathsf{C}(Y)}$. Why is this true? Well, first we have $f^*\mathcal{G}(U)=\mathcal{G}(f(U))$, for $U\subset X$ open, since $f$ is a homeomorphism. In other words, we have $f^*\mathcal{G}=(f^{-1})_*\mathcal{G}$, where $f^{-1}$ is the inverse function of $f$. Since this works for arbitrary $\mathcal{G}$, this implies $$\tag{3}\label{eq:3} f^*=(f^{-1})_*. $$ Also, the assignment $g\mapsto g_*$ is functorial. That is, the pushforward defines the action on morphisms of a functor $\mathsf{Sh}_\mathsf{C}(-):\mathsf{Top}\to\mathsf{CAT}$, where $\mathsf{CAT}$ is the category of locally small categories. It then follows from \eqref{eq:3} that $f^*f_*=1_{\mathsf{Sh}_\mathsf{C}(X)}$ and $f_*f^*=1_{\mathsf{Sh}_\mathsf{C}(Y)}$.

The equation \eqref{eq} can be used to prove equivalence $(2\Leftrightarrow 4)$ from the proposition. Observe that $f$ being a homeomorphism (i.e., a topological space isomorphism) implies that $f_*:\mathsf{Sh}_\mathsf{Ring}(X)\to\mathsf{Sh}_\mathsf{Ring}(Y)$ is a category isomorphism (from the functoriality of the assignment $g\mapsto g_*$). Thus, by \eqref{eq}, $f$ and $\varphi^\sharp$ are isos iff $f$ and $\varphi^\flat$ are isos.

Note that this argument does not use any particularity of the category $\mathsf{Ring}$ and it works for an arbitrary cocomplete category $\mathsf{C}$. The proposition is still true if stated for the “category of $\mathsf{C}$-sheafed spaces.” This category has as objects pairs $(X,\mathcal{F})$, where $X$ is a space and $\mathcal{F}\in\mathsf{Sh}_\mathsf{C}(X)$ is a $\mathsf{C}$-sheaf on $X$, and a morphism $(X,\mathcal{F})\to(Y,\mathcal{G})$ and composition of morphisms are defined just on the same fashion as in the case $\mathsf{C}=\mathsf{Ring}$. In particular, the category of $\mathsf{Ring}$-sheafed spaces coincides with the already familiar category of ringed spaces. The category of $\mathsf{C}$-sheafed spaces is addressed on EGA 0, 3.5.2., although there they do not give it a name.

Answer 2

Here is an easier answer. When I wrote the other answer I was a newbie in category theory and algebraic geometry.

The fact that [$1$ or $2$] $\Leftrightarrow$ [$3$ or $4$] follows from functoriality. We can consider a first type (resp., second type) category of sheafed spaces in which morphisms are given in the form $(f,\psi^\flat)$ (resp., in the form $(f,\psi^\sharp)$) and then define an isomorphism of categories from the first type to the second type by sending $(f,\psi^\flat)$ to $(f,\psi^\sharp)$. The fact that this assignment is indeed functorial is explained here.