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 $\beta = \alpha$.
Best Answer
I think you've seen why it ought to be the intersection, but for the sake of completeness I'll be especially painstaking about it.
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.