TITLE: Prove that a Boolean two-valued topos in which supports split is well-pointed
QUESTION [7 upvotes]: In Lawvere and Rosebrugh's Sets for Mathematics, they write

It is a theorem [MM92] that a topos is well-pointed if and only if it is Boolean, two-valued, and supports split.

[MM92] is a reference to Mac Lane and Moerdijk's Sheaves in Geometry and Logic. I have found the proof that a well-pointed topos is Boolean, two-valued and supports split (Propositions VI.7 and VI.8), but I cannot find the proof of the converse, there or anywhere else. Can anyone help me?

REPLY [6 votes]: Given a Boolean, two-valued topos in which supports split, we want to prove it's well-pointed, meaning that, if $f,g:A\to B$ are distinct, then there is a point $p:1\to A$ such that $fp\neq gp$.  Since $f$ and $g$ are distinct, their equalizer $e:E\to A$ is a proper subobject of $A$. (I'm ignoring the distinction between a subobject and a monomorphism representing it.) By "Boolean", $E$ has a  non-zero complement $c:C\to A$. Since $C\neq0$, the support of $C$ is a non-zero subobject of $1$. By "two-valued", the support of $C$ is all of $1$. By "supports split" we have a morphism $p:1\to C$. I claim this point in $C$, considered as a point in $A$ (strictly speaking, the point $cp:1\to A$) does the job. Indeed, if $fcp$ were equal to $gcp$, then $cp$ would factor through the equalizer $E$ as well as its complement $C$, hence would factor through 0. But then we'd have a morphism $1\to0$, and the topos would be trivial, contrary to the assumption $f\neq g$