I am reading this document:
http://www.tac.mta.ca/tac/reprints/articles/11/tr11.pdf
Compared with the ordinary definition of a topos, the axioms of ETCS do not assume the existence of a subobject classifier. In Theorem 5 of the above link, the author proved that every subobject has a characteristic map to the coproduct of two copies (denoted $2$) of the terminal object. I expect $2$ to be the subobject classifier. But I have trouble proving it. I want to prove that for every mono $a: A\to X$, the following diagram is a pullback (but actually, the word "pullback" does not appear in the whole article by Lawvere).
$$\require{AMScd} \begin{CD} A @>{!_A}>> 1;\\ @VVV @VVV \\ X @>{char_A}>> 2; \end{CD}$$
That is, given any object $T$ and arrow $t:T\to X$, if $char_A \circ t = i_1 \circ !_T$, then we want $t$ to factor through $A$.
For the construction of $char_A$, according to the proof of Theorem 5, I think it is:
$X \to A + A' \stackrel{(i_1,i_0)}\to 2$, where $(i_1, i_0)$ means the induced map from $A + A'$ ($A'$ is the complement of $A$), by the pair of maps $A \to 1 \stackrel{i1}\to 2$ and $A' \to 1 \stackrel{i0}\to 2$.
If I just play with the map $t: T\to X$ from an arbitrary object, I cannot see any hope of getting a map $T \to A$, since we do not have that $T \to A + A'$ implies $T$ factor through the injection from either $A$ or $A'$. By axiom 7, the coproduct is useful for a factorization only when we are considering maps from the terminal object.
Is there any way to prove that $2$ is the subobject classifier, using the definition "for each object $X$, for each subobject $A$, there is a unique characteristic map to $2$ such that the square diagram is a pullback"?
(If I am wrong in the sense that a subobject classifier does not exist in this setting, thank you for pointing it out.)
EDIT: I can prove that every element of $T$ compose with $t$ factors through $A$, but does it (and how does it) imply that there is a factorization $T\to A$ from the object $T$ itself?