Proof idea: The proof proceeds via construction of a contradiction if a subobject classifier $\Omega$ were to exist in $CAT$. I assume $CAT$ to be the meta-category of all categories with morphisms being functors. In specific, we show that $\Omega$ fails to classify a subcategory $\overline{T}$ of a category $T$ which is defined below.
Definition of $T$
$T$ consists of following data:
- Objects: $A$
- Morphisms: $1_A$, $\alpha_1$, $\alpha_2$
with composition defined as follows:
For any morphism $x$,
- $x\circ 1_A=1_A\circ x = x$
- $\alpha_1 \circ x = \alpha_1$
- $\alpha_2\circ x = \alpha_2$
Definition of $\overline{T}$
$A\in \overline{T}, 1_A\in \overline{T}, \alpha_1 \in \overline{T}$. Note that there exists an obvious inclusion map $\overline{t}:\overline{T}\hookrightarrow T$.
Firstly note that $\Omega$ results in following pull-back diagram: $$\require{AMScd} \begin{CD} \overline{T} @>{!_\overline{T}}>> 1\\ @V{\overline{t}}VV @VV{true}V \\ T @>>{\chi_\overline{(t)}}> \Omega \end{CD} $$ $$\tag*{1}$$
Lemma 0: $\chi_\overline{t}(\alpha_1)=1_{true}$.
This follows from commutativity of pull-back square in (1).
Lemma 1: $\chi_\overline{t}(\alpha_2)\ne 1_{true}$.
Again, this follows from the universal property of the pull-back square in (1), as otherwise replacing $\overline{t}$ with $1_T$ still results in a commutative square but $1_T$ cannot be factored through $\overline{t}$.
From definition of functor, it follows that $$\chi_\overline{t}(\alpha_1 \circ \alpha_2)=\chi_\overline{t}(\alpha_1)\circ \chi_\overline{t}(\alpha_2)$$
By applying Lemma 0 and definition of $\alpha_1 \circ \alpha_2=\alpha 1$, we get $$\chi_\overline{t}(\alpha_1)=1_{true}\circ \chi_\overline{t}(\alpha_2)$$
Simplfying a little bit more we find, $$\chi_\overline{t}(\alpha_1)=\chi_\overline{t}(\alpha_2)=1_{true}$$ Which contradicts Lemma 1. This means that $\chi_\overline{t}$ is not a functor, therefore does not exist in $CAT$. Q.E.D.
Dear reader,
Before finding this proof and writing this question, I did a modest search to find if there is already an explanation of why sub-object classifier must or cannot exist. But search only revealed proofs which were quite far from my current understanding i.e. inaccessible for me. So, I thought why not at least try to see if there is an elementary proof?
I am just posting this so that it may be useful to others and also to check if I have got the argument sound.
I welcome any advice/comments that may be helpful.
Thank you for your time and effort.
Yes, that looks good. More abstractly the problem is that if $A\to B$ is the pullback of some map $\top:1\to C$ in CAT, then $A$ satisfies the "two-out-of-three" property: given $f:x\to y$ and $g:y\to z$ in $B$, if any two of $f,g,$ and $gf$ are in $A$, then so is the third. This is because if any two of $f,g,$ and $gf$ are sent to an identity in $C$, then so is the third. Your example illustrates this general problem, since $\bar T\to T$ does not satisfy two-out-of-three: $\alpha_1$ and $\alpha_1\circ \alpha_2$ are in $\bar T$, but $\alpha_2$ is not.