# Subobject classifier of disjoint sum of subobjects

Question

Let $$\mathcal{E}$$ be a topos with subobject classifier $$1 \xrightarrow{t} \Omega$$. Consider the subobject $$1 \xrightarrow{\langle t, t \rangle} \Omega\times\Omega$$ and call its subobject classifier $$F$$. Let $$A, B$$ be subobjects of $$X$$ classified by $$\phi, \psi$$, then what is the subobject of $$X$$ classified by $$X \xrightarrow{\langle \phi, \psi \rangle} \Omega \times \Omega \xrightarrow{F} \Omega$$?

My work

I think it should be the map $$A \sqcup B \rightarrow X$$ induced by $$A \to X$$ and $$B \to X$$ since that is the only one I could think of. Proving $$A \sqcup B$$ is the pullback of $$F \circ \langle \phi, \psi \rangle$$ and $$t$$ is equivalent to proving it is the pullback of $$\langle \phi, \psi \rangle$$ and $$\langle t, t \rangle$$ by the pullback lemma.

Suppose we have an object $$Y$$ and an arrow $$f: Y \to X$$ such that $$\langle \phi, \psi \rangle \circ f = Y \to 1 \xrightarrow{\langle t, t \rangle} \Omega \times \Omega$$. By composing with the projection of $$\Omega \times \Omega \to \Omega$$ we find that $$\phi \circ f = Y \to 1 \xrightarrow{t} \Omega$$, so we obtain a unique arrow $$\sigma: Y \to A$$ such that $$Y \xrightarrow{\sigma} A \to X = f$$. Similarly we obtain a unique $$\tau: Y \to B$$ such that $$Y \xrightarrow{\tau} B \to X = f$$. Composing $$\sigma$$ with the inclusion $$i_1$$ of $$A$$ into $$A \sqcup B$$ yields a suitable morphism, but composing $$\tau$$ with the inclusion $$i_2$$ of $$B$$ into $$A \sqcup B$$ as well. Assuming that works, I also cannot prove that this is the only morphism $$Y \to A \sqcup B$$ that gives $$f$$ composed with $$A \sqcup B \to X$$.

Picture for reference:

Edit: as Malice Vidrine pointed out, it should be the pullback $$P$$ of $$A \to X \xleftarrow{} B$$. The morphisms $$\sigma$$ and $$\tau$$ induce a morphism $$\alpha: Y \to P$$. We have that the composite of $$\alpha$$ and $$P \to X$$ is equal to $$Y \xrightarrow{\sigma} A \to X = Y \xrightarrow{\tau} B \to X = f$$. If some $$\beta$$ satisfied this as well, then $$Y \xrightarrow{\beta} P \to A = \sigma$$ and $$Y \xrightarrow{\beta} P \to B = \tau$$ by the universal property of the pullbacks of $$\phi, t$$ and $$\psi, t$$, but then by the universal property of $$A \to P \xleftarrow{} B$$, we have that $$\beta = \alpha$$.

• It should be the intersection of subobjects, not the join. – Malice Vidrine Jan 8 at 22:12
• @MaliceVidrine Thanks! You should make an answer of that comment. – Pel de Pinda Jan 8 at 22:22
• If no one beats me to it, I'll try when I'm back at my computer :) – Malice Vidrine Jan 8 at 22:39

Let's just use $$i:E\rightarrowtail X$$ for the subobject classified by $$F\circ\langle\phi,\psi\rangle$$, and $$i_A,i_B$$ be the inclusion morphisms of $$A,B$$, respectively, into $$X$$.
By the lefthand commutative square in the above diagram, we have that $$\phi\circ i=\top\circ !_E$$ (so $$i$$ factors through $$i_A$$) and $$\psi\circ i=\top\circ !_E$$ (so $$i$$ factors through $$i_B$$). It follows from this (along with the fact that $$i$$ is a monomorphism) that $$E$$ is a subobject of both $$A$$ and $$B$$, and this is our first clue that we're looking at what ought to be an intersection. Let's denote by $$j_A,j_B$$ the respective inclusions of $$E$$ into $$A,B$$; note, in particular, that $$i_Aj_A=i=i_Bj_B$$.
Now let $$f:Q\to A$$ and $$g:Q\to B$$ be morphisms such that $$i_Af=i_Bg$$; we wish to show that there is a unique map $$h:Q\to E$$ with $$j_A\circ h=f$$ and $$j_B\circ h=g$$; that is, we want to show that $$E$$ is the pullback of $$i_A$$ and $$i_B$$.
Denote by $$\alpha$$ the morphism $$i_Af=i_Bg:Q\to X$$. Then $$\phi\circ\alpha=\phi\circ i_A\circ f=\top\circ !_A\circ f=\top\circ!_Q$$ and $$\psi\circ\alpha=\psi\circ i_B\circ g=\top\circ !_B\circ g=\top\circ!_Q,$$ meaning that $$\langle\phi,\psi\rangle\circ\alpha=\langle\top,\top\rangle\circ !_Q$$. Because that left hand square is a pullback, there is a unique morphism $$h:Q\to E$$ with $$i\circ h=\alpha$$. This equality gives us both $$i_A\circ j_A\circ h=i\circ h=\alpha=i_A\circ f$$ from which we have $$j_Ah=f$$ because $$i_A$$ is a monomorphism, and $$i_B\circ j_B\circ h=i\circ h=\alpha=i_B\circ g$$ from which we have $$j_B\circ h=g$$.
Thus we've shown that $$E$$ satisfies the universal property of the intersection of $$A$$ and $$B$$, using only that the squares in the pictured diagram are pullbacks.