I'm reading Awodey, and I'm confused about this part:
Firstly, the map $\top!$ is only defined on $U$, not on the whole $A$. But it must be defined on $A$ (equalizers defined for maps out of $A$). Is it supposed to be defined on the whole $A$ by the same formula?
Secondly, what is $1$? Is it the terminal object in the category of sets? That is, is $1=\{\ast\}$? Further, what is the map $\top$? Is it $\ast \mapsto \top$?
Finally, the original statement is that $U$ together with the inclusion map $U\to A$ is an equalizer (i.e., the inclusion followed by $\top!$ and the inclusion followed by $\chi_U$ are equal, and given any set $Z$ and a map $h: Z\to A$ with $\top!\circ h=\chi_U\circ h$ there is a unique $u:Z\to U$ that makes the triangle commute). I can see that the compositions are equal. But the universal property is unclear. Suppose $fh=fg$. How to define the map $u: Z\to U$?
The first two questions seem to have been answered by Max in the comments: The $U$ is a typo; it should be $A,$ and yes, $1$ is the terminal object and the map $\top$ is the map from $1$ that has value $\top$ (i.e. the 'element' $\top$ in the category theory sense.)
For the third, note that $\top!\circ h=\chi_U\circ h$ holds if and only if the image of $h$ is contained in $U$. Thus the map $u: Z\to U$ that you are looking for is simply $h,$ viewed as a function with codomain $U,$ rather than with codomain $A.$