"False" arrow in an elementary topos

Question

In "Topoi: The Categorial Analysis of Logic" by R. Goldblatt (page 117) the "false" arrow is defined based on the comutative diagram: $\require{AMScd}$ \begin{CD} 0 @>0_1>> 1\\ @V ! V V @VV {\rm false} V\\ 1 @>>{\rm true}> \Omega \end{CD} Now, the arrow into the final object is unique, so it seems that $0_1$ and the arrow denoted by $!$ must be the same. But this would make ${\rm true}={\rm false}$, which makes no sense. It looks like I am missing something very basic. Where am I mistaken?

Answer 1

The arrow ${\rm false}$ is not defined by this square merely commuting, but by it being a pullback square. So, you are correct that $0_1$ and $!$ are the same map, and so if you replaced ${\rm false}$ with ${\rm true}$ then the diagram would still commute. However, to conclude that ${\rm true}={\rm false}$, you would need to know the diagram remains a pullback square, not just that it commutes.

(For some context, recall that $1\stackrel{{\rm true}}\to\Omega$ has the universal property that for any objects $A$ and $B$ and any monic morphism $f:A\to B$, there is a unique morphism $g:B\to \Omega$ such that $$\begin{CD} A @>f>> B\\ @V ! V V @VVgV\\ 1 @>>{\rm true}> \Omega \end{CD}$$ is a pullback square. So, this universal property is being applied to the monomorphism $0_1:0\to 1$ (which is just the unique morphism from $0$ to $1$), with the unique corresponding morphism $1\to\Omega$ being named ${\rm false}$.)