Subobject Classifier of a Topos is Injective

Question

Let $\mathcal{E}$ be a topos with subobject classifier $1\overset{t}{\rightarrow}\Omega$, then I want to show that $\Omega$ is injective.

So let $f:A\rightarrow\Omega$ and $g:A\rightarrow B$ be two maps in $\mathcal{E}$ with $g$ monic. Then I want to show that there exists a map $h:B\rightarrow\Omega$ such that $h\circ g = f$. Notice that by the property of a subobject classifier we have that there exists a unique map $\phi:B\rightarrow\Omega$ such that there is a pullback square $\require{AMScd}$ \begin{CD} A @>{\psi}>> 1\\ @VV{g}V @VV{t}V\\ B @>{\phi}>> \Omega \end{CD} Now I claim that $\phi$ does the job for $h$. But I don't see why $\phi\circ g= f$ (so maybe my claim is wrong). Any help would be appreciated!

Answer 1

Thinking about maps to $\Omega$ as subobjects, this is just saying that if you have a subobject $C\subseteq A$, then there is a subobject $D\subseteq B$ such that $C=A\cap D$ as subobjects of $B$. How do you find such a $D$? Well, you can just take $D=C$!

To make this precise, let $i:C\to A$ be the subobject classified by $f$. Then $gi:C\to B$ is monic since $i$ and $g$ are, so it is classified by a map $h:B\to\Omega$. Now in the diagram $$\require{AMScd} \begin{CD} C @>{1}>> C @>{}>> 1\\ @V{i}VV @VV{gi}V @VV{t}V\\ A @>{g}>> B @>{h}>> \Omega \end{CD} $$ the right square is a pullback by definition of $h$ and the left square is a pullback since $g$ is monic. This implies the outer rectangle is a pullback, i.e. that $hg$ classifies the subobject $i:C\to A$, so $hg=f$.

Answer 2

Eric's answer is spot on.

Here is an alternative method in which you might also be interested. Employing the internal language of the topos, we may reason in a naive element-based language similar to the ordinal mathematical language, as long as we take to restrict to intuitionistic logic (that is, not use the law of excluded middle). The question then becomes this:

Given an injective map $g : A \to B$ between sets and a map $f : A \to \Omega$, where $\Omega$ is the powerset of $1 = \{\heartsuit\}$, why is there an extension $g : B \to \Omega$?

And the answer is: We may set $g(y) := \{ p \in 1 \,|\, y \in A \Rightarrow 1 \in f(y) \}$. For $y \in A$, we have $g(y) = f(y)$, hence this map $g$ is indeed an extension.