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An isomorphism of ringed spaces is a morphism of ringed spaces

$(f,\theta) :(X, \mathcal{O}_{X})\rightarrow (Y,\mathcal{O}_{Y})$

$(f:X\rightarrow Y$ continuous, $\theta: \mathcal{O}_{Y}\rightarrow f_{*}\mathcal{O}_{X}$ a morphism of sheaves of rings)

which admits a two sided inverse. In particular, $f$ is a homeomorphism and $\theta$ is an isomorphism of sheaves.

Is the converse also true? That is, given a morphism of ringed spaces $(f,\theta)$ such that $f$ is a homeomorphism and $\theta$ and isomorphism of sheaves of rings, is this also an isomorphism of ringed spaces?

The answer seems to be yes. However, I do not see why. Here are some of my thoughts:

Let $g$ be the inverse of $f$ and $\tau$ be the inverse of $\theta$. More precisely, $\tau$ is a morphism of sheaves of the form $f_{*}\mathcal{O}_{X}\rightarrow \mathcal{O}_{Y}=f_{*}g_{*}\mathcal{O}_{Y}$. My question would be positively answered, if $\tau$ were of the form $f_{*}\rho$ for some morphism of sheaves $\rho:\mathcal{O}_{X}\rightarrow g_{*}\mathcal{O}_{Y}$, so for my purposes it is a convenient hope that $f_{*}$ is a full functor. Since it is a right adjoint, this amounts to saying that the counit map $f^{-1}f_{*} F\Rightarrow F$ is a split monomorphism for any sheaf $F$, a statement I am neither able to prove nor to disprove. So this is where a second question sneaks in: Is $f_{*}$ full?

Thanks in advance!

user520682
  • 427

2 Answers2

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You can define $\rho\colon \mathcal{O}_{X}\to g_{*}\mathcal{O}_{Y}$ directly by hand. Let $U\subseteq X$ be an open set. Then $V=f(U)\subseteq Y$ is open, so we have an isomorphism $$ \theta|_{V}\colon \mathcal{O}_{Y}(V)\to \mathcal{O}_{X}(f^{-1}(V))=\mathcal{O}_{X}(U) $$ Define then the component of $\rho$ at $U$ to be its inverse: $$ \rho|_{U}=(\theta|_{V})^{-1}\colon \mathcal{O}_{X}(U)\to \mathcal{O}_{Y}(g^{-1}(U))=\mathcal{O}_{Y}(V) $$ Applying the functor $f_{*}$ we obtain the natural transformation of sheaves on $Y$ whose component at $V$ is given by $$\tau|_{V}=\rho|_{f^{-1}(V)}=\rho|_{U}=(\theta|_{V})^{-1}$$ So the inverse $\tau$ is indeed of the form $f_{*}\rho$.

Pedro
  • 4,994
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If $\theta$ is an isomorphism, then so is $g_{*}\theta$ and the inverse map $(g_{*}\theta)^{-1}:\mathcal{O}_{X}\rightarrow g_{*}\mathcal{O}_{Y}$ is a morphism of sheaves of rings such that $f{*}((g_{*}\theta)^{-1})=\theta^{-1}$ by functoriality and the fact that $f$ and $g$ are mutually inverse.

user520682
  • 427