TITLE: Subobject Classifier of a Topos is Injective
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: 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.