Why is $[\top,\bot]$ the character of the coproduct inclusion $i_1$? $\require{AMScd}$In an elementary topos $\mathbf{C}$, why should $m=[\top,\bot]$ be the character morphism of the first coproduct inclusion $i_0:\mathbf{1}\to\mathbf{1}+\mathbf{1}$?
I have tried to show that if we take $k:P\to\mathbf{1}+\mathbf{1}$ as the pullback of $\top$ and $m$
$$ \begin{CD} P @>k>> \mathbf{1}+\mathbf{1} \\ @V!_PVV @VVmV \\ \mathbf{1} @>>\top> \Omega\end{CD}$$
then $P$ is a terminal object and thus $P\cong\mathbf{1}$ so that
$$ \begin{CD} \mathbf{1} @>i_0>> \mathbf{1}+\mathbf{1} \\ @V=VV @VVmV \\ \mathbf{1} @>>\top> \Omega\end{CD}$$
is also a pullback.
This is motivated by my previous question Proof that $[\top,\bot]$ is monic and the first answer that was given.
Thanks!
 A: The idea of using the universal property of the pullback to show that $P$ in your diagram must be $1$ is a good one!
To show that there's a (unique) arrow from my favorite object $X$ into $P$, we have to show that there's a (unique) pair of arrows from $X \to 1$ and $X \to 1+1$ making the diagram commute.
Of course, we're halfway there out of the gate, since we know there's a unique arrow $X \to 1$. For the other arrow, since $m = [\top, \bot]$ it's easy to see that the arrow $i_0 \ \circ \ ! : X \to 1+1$ makes the diagram commute.  This also gives us $k = i_0$ when we set $X = P = 1$.
As an only-slightly-tricky exercise, can you show that $i_0 \ \circ \ !$ was the only choice of $X \to 1+1$ making the diagram commute, so we really do get uniqueness?

Edit:
We know that toposes are extensive, so if $f : X \to 1+1$ we know that $X$ is actually isomorphic to $X_0 + X_1$ where $f \restriction X_j = i_j \ \circ \ !$.
Then if $f$ makes the diagram commute, we know that
$$X_j \hookrightarrow X \overset{f}{\to} 1+1 \overset{[\top,\bot]}{\longrightarrow} \Omega = \top$$
for both $X_j$. Of course, since $\top \neq \bot$, this means $X_1 \cong 0$, $X \cong X_0$, and $f = i_0 \ \circ \ !$.

I hope this helps ^_^
