I am reading the following document about Lawvere's ETCS:
http://www.tac.mta.ca/tac/reprints/articles/11/tr11.pdf
My question is about Theorem 6 in this document. The aim is to prove an equivalence relation is an equalizer. On page 29, the author says:
Finally we can assert our theorem (I think the $a_0q = a_1g$ in the link is a typo.) $a_0q = a_1q ⇔ a_0 ≡ a_1$.
And finished the proof.
That is, the author proves for every element $\langle a_0,a_1\rangle$ from the terminal object $1$, we have the dashed map as in the picture. But to prove the equalizer result, we need that we still have the factorization when $1$ is replaced by any arbitrary object. I know that $1$ is the generator in ETCS, and have trouble working out the proof that the existence of factorization for $1$ implies the existence of factorization for any $T$. I tried taking the elements of $T$, and unable to construct a factorization from $T$ from the factorization for its elements.
Any help, please? Thank you!
Note: If someone is reading the link above, then please note that the composition is written as: if $f: A \to B,g: B\to C$, then the author will write $fg$ to denote the composition of $f$ of $g$, rather than $g \circ f$.