Does the category of measurable spaces have a strong subobject classifier (specifically $2 = (\{0,1\}, \{\varnothing, \{0,1\}\})$?
I would think the situation could be analogous to $\mathsf{Top}$, which is also topologically concrete, but for some reason I can't find anything on this topic just googling (including a characterization of strong monos).
Best Answer
It's true, although maybe somewhat tedious to prove directly from scratch. I'll prove some intermediate results first (also for easy reference). I claim no originality, but couldn't find a reference anywhere.
We denote by $1_A : X\to 2$ the indicator function of $A\subseteq X$.
Lemma: In $\mathsf{Meas}$ (the category of measurable spaces and maps) a morphism $e : S\to E$ is epi iff it's surjective.
Proof: The forgetful functor into $\mathsf{Set}$ is faithful, hence surjective implies epi. Conversely, let $e$ be epi. Since
$$1_{e(S)} \circ e = 1 = 1\circ e,$$
and any map into $2$ is automatically measurable, we have $1_{e(S)} = 1$, i.e. $e$ is surjective. $\square$
Lemma: In $\mathsf{Meas}$ a morphism $i : S \to X$ is a strong mono iff it's a subspace embedding, i.e. $i$ is injective and the $\sigma$-algebra $\mathcal F_S$ on $S$ is exactly the $\sigma$-algebra generated by $i$
$$\mathcal F_S = \sigma(i) = \{i^{-1}(A) : A\in \mathcal F_X\}.$$
Proof: Since $\mathsf{Meas}$ is topologically concrete is (bi-) complete, hence extremal = strong. Hence, we can work with extremal monos.
"$\Leftarrow$": Suppose $i : S\to X$ is a subspace embedding, which factors as $i = \phi \circ e$ in $\mathsf{Meas}$ with $e$ an epi. We need to show $e$ is iso. Since $i$ is injective, so is $e$ and hence $e$ is bijective. We need to show that $e^{-1}$ is measurable, i.e. we claim
$$e(i^{-1}(A)) = (e^{-1})^{-1}(i^{-1}(A)) \in \mathcal F_E$$
for any $A\in \mathcal F_X$. Indeed, $e(i^{-1}(A)) = e \circ e^{-1}(\phi^{-1}(A)) = \phi^{-1}(A)\in \mathcal F_E$, since $\phi$ is measurable. Thus, $e$ is an iso and in conclusion $i$ is an extremal mono.
"$\Rightarrow$": Suppose $i$ is an extremal mono. Then $e : S\to i(S), x\mapsto i(x)$ is a well-defined surjection. We consider $i(S)$ as the image of $i$ equipped with subspace $\sigma$-algebra
$$\mathcal F_{i(S)} = \{i(S) \cap A : A\in \mathcal F_X\}.$$
Then $e^{-1}(i(S)\cap A) = i^{-1}(i(S)\cap A)\in \mathcal F_S$, so $e$ is measurable and hence an epi in $\mathsf{Meas}$.
Further, $i = e\circ \phi$, where $\phi : i(S)\to X, x\mapsto x$ is (measurable) the subspace inclusion. By assumption $e$ is iso. In particular,
$$\mathcal F_S = \sigma(e) = \{i^{-1}(A\cap i(S)) : A\in \mathcal F_X\} = \sigma(i),$$
where we used th injectivity of $i$ in the last step. Hence, $i$ is subspace embedding. $\square$
Proposition: $2$ is a strong subobject classifier in $\mathsf{Meas}$.
Proof: It's enough to show that it classifies subspace embeddings. Let $i : S \to X$ be a subspace embedding and $\top : 1\to 2$ be the map which maps the unique element in the singleton space / terminal object to $1 \in 2$.
Consider the diagram
The square commutes, since
$$1_{i(S)} \circ i = 1 = \top \circ !.$$
Now given $q$, such that
$$1_{i(S)} \circ q = 1$$
we have $q(Q) \subseteq i(S)$. Hence, for every $x\in Q$ there exists a $y\in M$, such that $i(y) = q(x)$. It mus be unique, since $i$ is injective. Hence, the map
$$u : Q\to M, x\mapsto y\text{ s.t. }i(y) = q(x)$$
is well-defined and satisfies $i\circ u = q$. Further, if there is another $u'$ with this property, then $i(u(x)) = q(x) = i(u'(x))$ for all $x\in Q$, hence $u = u'$ because $i$ is injective.
Therefore, the square is a pullback. We conclude that $2$ is classifier of strong subobjects. $\square$