# When is the inverse image of the structure sheaf the structure sheaf?

For which morphisms of schemes $f : X\rightarrow Y$ do we have $f^{-1}\mathcal{O}_Y = \mathcal{O}_X$? (note that I don't mean $f^*$, I really just mean the basic inverse image sheaf as a sheaf of rings/abelian groups/sets)

This is obviously true if $f$ is an open immersion, and intuitively I feel like this is the only time when it is true. Is this correct? (If not, is it possible to completely characterize the morphisms $f$ for which this is true?)

I was led to think about this when I read here that for a morphism of stacks $f : \mathcal{X}\rightarrow\mathcal{Y}$ (say over $(\textbf{Sch}/S)_{et}$, there is a canonical identification $f^{-1}\mathcal{O}_\mathcal{Y} = \mathcal{O}_\mathcal{X}$, which I feel is somewhat strange. I suppose this is just a peculiarity of working with big sites?

• Let $X$ be a scheme and $V$ be a sheaf of set on the underlying topological space of $X$. Then the étale space of $V$ endowed with the pullback of the structural sheaf of $X$ is a scheme satisfying this property (it is a scheme because it is locally isomorphic to $X$, it might not be a separated scheme in general). I believe those are the only ones. – Simon Henry Nov 23 '17 at 19:16
• But note how they are just the "sheaves of set over the small Zariski topos", In your case you just look at a different topos (and sheaves of groupoids instead of sheaves of sets) but the situation is not that different. – Simon Henry Nov 23 '17 at 19:18

Here is a trivial lemma.

Lemma. Let $f \colon X \to Y$ be a morphism of schemes. Then the natural map $f^{-1}\mathcal O_Y \to \mathcal O_X$ is an isomorphism if and only if for each $x \in X$ the natural map $\mathcal O_{Y,f(x)} \to \mathcal O_{X,x}$ is an isomorphism.

Proof. A morphism of sheaves on $X$ is an isomorphism if and only if it is so at the stalk of every $x \in X$. The stalk of a sheaf $\mathscr F$ at $\iota \colon \{x\} \to X$ is given by $\iota^{-1}\mathscr F$, so $(f^{-1}\mathcal O_Y)_x = \mathcal O_{Y,f(x)}$ since $\iota^{-1}f^{-1} = (f \circ \iota)^{-1}$. $\square$

Under reasonable finiteness assumptions, this gives a local isomorphism as Simon Henry suggested:

Lemma. Let $f \colon X \to Y$ be a morphism of schemes that is locally of finite presentation. Then $f^{-1}\mathcal O_Y = \mathcal O_X$ is and only if for every point $x \in X$, there exists an open neighbourhood $U \subseteq X$ of $x$ and an open neighbourhood $V \subseteq Y$ of $f(x)$ such that $f$ induces an isomorphism $U \stackrel\sim\to V$.

Proof. Suppose $f^{-1}\mathcal O_Y = \mathcal O_X$, and let $x \in X$. By the lemma above, we get $\mathcal O_{Y,f(x)} \stackrel\sim\to \mathcal O_{X,x}$. Let $V \subseteq Y$ be an affine open neighbourhood of $f(x)$ and $U \subseteq f^{-1}(V)$ an affine open neighbourhood of $x$. If $V = \operatorname{Spec} A$ and $U = \operatorname{Spec} B$, then the map $g \colon A \to B$ is of finite presentation (Tag 01TQ).

Letting $\mathfrak p \subseteq A$ and $\mathfrak q \subseteq B$ be the primes corresponding to $f(x) \in V$ and $x \in U$ respectively, we conclude that $g^{-1}(\mathfrak q) = \mathfrak p$, and the map $A_\mathfrak p \to B_\mathfrak q$ is an isomorphism. By Tag 00QS, this implies that there exist $a \in A\setminus \mathfrak p$ and $b \in B\setminus \mathfrak q$ such that $A_a \cong B_b$.

Replacing $U$ by $D(b)$ and $V$ by $D(a)$ gives open neighbourhoods $U \subseteq X$ of $x$ and $V \subseteq Y$ of $f(x)$ such that $f$ induces an isomorphism $U \stackrel \sim \to V$. The converse is trivial. $\square$

Examples include open immersions, disjoint unions, but also the map from a line with double origin to the line with a single origin.

However, if we drop the finite presentation assumption, we get all sorts of examples:

Example. Let $Y$ be any scheme, and consider the map $$X = \coprod_{y \in Y} \operatorname{Spec} \mathcal O_{Y,y} \to Y.$$ For a point $x \in X$ in the component $\operatorname{Spec} \mathcal O_{Y,y}$, we get an isomorphism $(\mathcal O_{Y,y})_x \to \mathcal O_{Y,f(x)}$, as can be seen easily by passing to an affine open neighbourhood of $y$ in $Y$ and using standard properties of localisation of rings.