Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 75 / 4

The defining square for the subobject classifier is a pullback in a category:The inverse image functor of a geometric morphism preserves this square and the terminal object. By definition, the inverse-image truth arrow $f^{\ast }{\top }_{\mathcal{E}}$ is itself the pullback of ${\top }_{\mathcal{F}}$ along $\bar{\lambda }$. Composing these two pullback squares shows that $f^{\ast }A\hookrightarrow f^{\ast }X$ is the truth pullback along $\bar{\lambda }\circ f^{\ast }a$. Uniqueness of the characteristic map gives

Take two subobjects $A,B\hookrightarrow X$ whose inverse images coincide. Their characteristic maps $a,b:X\to {\Omega }_{\mathcal{E}}$ satisfy $\bar{\lambda }{f}^{\ast }a=\bar{\lambda }{f}^{\ast }b$ by part (i). Under the adjunction $f^{\ast }⊣f_{\ast }$, these maps transpose to $\lambda a$ and $\lambda b$, respectively, by naturality of the unit. If $\lambda$ is a monomorphism, then $a=b$ and hence $A=B$. Thus $f^{\ast }$ reflects equality of subobjects.

For parallel arrows $r,s:X\rightrightarrows Y$ with $f^{\ast }r=f^{\ast }s$, their equalizer $K\hookrightarrow X$ pulls back to all of $f^{\ast }X$. Reflection of equality with the whole subobject gives $K=X$, so $r=s$. Therefore the inverse image is faithful, and the geometric morphism is surjective. This is the surjectivity detected by the subobject classifier criterion. For completeness, faithfulness is equivalent here to conservativity: if $f^{\ast }h$ is invertible, the image inclusion of $h$ and its kernel-pair diagonal become whole subobjects; reflection makes $h$ both a regular epimorphism and monic, hence invertible in the topos.

Suppose $f$ is a surjective geometric morphism, so $f^{\ast }$ is faithful. First, if a monomorphism $m:S\hookrightarrow X$ becomes an isomorphism, then its characteristic map ${\chi }_m$ and the constantly true map become equal after $f^{\ast }$: by part (i), the pulled-back subobject is the whole object, and equivalently $f^{\ast }{\chi }_m=f^{\ast }{\top }_X$ because $f^{\ast }m$ is epic and these maps agree after precomposition with it. Faithfulness gives ${\chi }_m={\top }_X$, so $m$ was already invertible.

Now let $a,b:X\to {\Omega }_{\mathcal{E}}$ satisfy $\lambda a=\lambda b$. Adjunction and part (i) identify their pulled-back subobjects $f^{\ast }A=f^{\ast }B$. Each inclusion $A\cap B\hookrightarrow A,B$ consequently becomes an isomorphism. By the preceding argument it already is an isomorphism, so $A=B$ and $a=b$. This works for every test object $X$, proving