I am currently reading Goldblatt's Topoi, A categorial analysis of logic, and on p.117 there is exercise 3 where we have to show that $false_a$ is the character for $0_a$. There is a hint recommending to use the PBL. So far I've come to the following diagram :
$\require{AMScd}$ \begin{CD} 0 @>!>> 1 \\ @V 0_a V V @VV true V\\ a @>>false_a> \Omega \\ @V I_a V V @VV {1_\Omega} V \\ 1 @> false >> \Omega \end{CD}
Clearly the outer rectangle is a pullback, and in order to conclude with the PBL, I want to show that the down square is also a pullback, but I'm stuck there. Does anyone have a hint as to why this is the case ?
Thanks
Consider a square whose two paths are $0\to 0\to 1$ and $0\overset{0_a}{\to} a\to 1$ (and picture the first of these as the left and bottom sides of a square). I claim that this is a pullback, since if we are given $f:x\to 0$ and $g:x\to a$ with $$x\overset{f}{\to}0\to 1=x\overset{g}{\to}a\to 1,$$ the existence of a morphism $x\to 0$ immediately means that $x$ is also initial, and there is a unique morphism $x\to 0$ filling in the pullback diagram for very silly reasons.
Combine this with the pullback square that defines $false$ and you're done.