Surjection Vs Surjective geometric morphism

Question

Is it true that a map between ${\bf T1}$ topological spaces $f:X \to Y$ is surjective iff the induced geometric morphism $f:Sh(Y) \to Sh(X)$ is a surjection (i.e. its inverse image part $f^*$ is faithful)?

In "Sheaves in Geometry and Logic" a proof is given, but the the "if" part leaves me a bit unsatisfied (and, btw, the "only if" part holds for any topological space). Thanks in advance!

Answer 1

I think you want $Sh(Y) \rightarrow Sh(X)$ to be an injection not a surjection, otherwise take $Y = pt$, we're saying all the sheaves on everything are constant!

Let's consider sheaves of abelian groups, just to make faithfulness easier to think about.

Well, if $X \rightarrow Y$ is not surjective, take a skyscraper on a point not in image, it should pull back to zero. (this is where you use $T_0$, btw, to conclude all the stalks away from the point are 0).

If $X \rightarrow Y$ is surjective, then since the stalk of $f^* \mathcal{F}$ at $x$ is the stalk of $\mathcal{F}$ at $f(x)$, if $f^* \mathcal{F}$ is 0, all the stalks of $\mathcal{F}$ are 0 i.e. its zero.

Answer 2

Here is some elucidation on Mac Lane and Moerdijk's proof.

"Suppose a point $y \in Y$ is not in the image of $f$." This makes $f^{-1}(Y - \{y\}) = f^{-1}(Y) = X$, since all points of $X$ are mapped into $Y - \{y\}$.

"Since $Y$ is assumed to be $T_1$, $y$ is a closed point, so $Y - \{y\}$ is open." That's because all points are closed in $T_1$ spaces.

"Therefore $Y - \{y\} = Y$, since $f^{-1}$ is faithful." As @Zhen_Lin mentioned in the comments, we actually need the extra information that $f^{-1}$ is conservative to reach this conclusion, or at least that $f^{-1}$ is injective on subobject lattices, so you were right to be skeptical. While this is proved in Lemma VII.4.3 of the book using subobject classifiers, it is much easier to observe that faithful functors reflect monomorphisms and epimorphisms, and so the fact that $f^*$ is faithful and that a morphism in a topos is an isomorphism if and only if it is monic and epic implies that $f^*$ is conservative (and hence so is its restriction $f^{-1}$).

The strategy for an arbitrary space is to consider the stalks across all points, showing that for $\alpha,\beta:E \to F$ in $\mathrm{Sh}(Y)$ with $f^*(\alpha) = f^*(\beta)$ we have $\alpha_y = \beta_y$ for all points $y$ and hence $\alpha = \beta$, because everything is determined by the values on stalks. As you and they point out, holds for any space.