Counterexample for statement: $2$ is always a valid subobject classifier

Question

In the category $\mathbf{Sets}$, the subobject classifier is $2$, i.e. a two-element set. I understand that, but I don't understand why $\Omega$ is not $2$-like for some of the other categories.

By "$2$-like object" I mean such object $A$, so that there are exactly two different morphisms $1 \rightarrow A$.

Please walk me through a counterexample in the following format:

1. Pick a category $C$ that has a subobject classifier which is not $2$-like.
2. Forget about the original $\Omega$ and assume that the subobject classifier $2$-like.
3. Show that the definition of a subobject classifier does not hold (e.g. you can't form the pullback square).

I don't want to ask for a specific counterexample category, because I don't know in which category the counterexample will be the easiest to show. If $\mathbf{Sets \times Sets}$ or $\mathbf{Sets^2}$ are OK, then please use them.

Answer 1

$\text{Set}^2$ will work fine, but because of the multiple appearances of the number $2$ it will be less confusing to work with $\text{Set}^n$ for an arbitrary positive integer $n$ (we could more generally consider $\text{Set}^I$ for a possibly infinite index set $I$).

The subobject classifier is the object $2$ in the sense that it is the coproduct $1 \sqcup 1$; explicitly it is the object $(2, 2, 2, \dots)$. But there are $2^n$ morphisms $1 \to 2$, since we can pick a different one in each coordinate, so the subobject classifier is not "$2$-like" according to your definition.

In general, by the universal property of the subobject classifier, morphisms $1 \to \Omega$ are naturally in bijection with subobjects of $1$, that is, with equivalence classes of monomorphisms $m \hookrightarrow 1$. The subobjects of the terminal object are called subterminal objects, and it's not hard to see that in $\text{Set}^n$ the subterminal objects are the objects which are either $0$ (the empty set) or $1$ (a $1$-element set) in each coordinate; in particular there are (up to equivalence) $2^n$ of them, which matches the count above. This is another way to see that the subobject classifier can't be "$2$-like."