Join in Lattice of Subobjects In an elementary topos the join $A \vee B$ of two subjects $A \to X$ and $B \to X$ is defined to be the image of the induced morphism $A \sqcup B \to X$. For sets it holds, that this is the same as the pushout of $A$ and $B$ along their intersection $A \cap B = A \wedge B = A \times_X B$. Does this hold in general?
Here the image of a morphism $f\colon M \to X$ is defined to be the equalizer of the two inclusions $X \to X \cup_M X$.
 A: Yes, it holds in general.
A coherent category is one with finite limits, where every map has a pullback-stable image, and where for all objects $A$, $Sub(A)$ is a bounded lattice, and where pullbacks preserve finite joins. Every topos is a coherent category.
It can be shown that in any coherent category, the property you stated holds. In fact, in general, if you have subobjects $S_1, \ldots, S_n \subseteq X$, then $S_1 \lor \cdots \lor S_n$ is the colimit of the obvious diagram involving the $S_i$s and the inclusion maps $S_i \land S_j \to S_i$.
To prove this, suppose WLOG that $X = S_1 \lor \cdots \lor S_n$. Suppose we have functions $f_i : S_i \to Y$ such that $f_i |_{S_i \land S_j} = f_j |_{S_i \land S_j}$.
For intuition’s sake, in set theory, we would build the function $X \to Y$ by defining a subset of $X \times Y$ - specifically, the subset $W = \{(x, y) \mid \exists i (x \in S_i \land y = f_i(x))\}$.
Now consider that these $f_i$, together with inclusion maps, give rise to monics $g_i : S_i \to X \times Y$. Take their join $g : W \to X \times Y$.
In set theory, the next step would be showing that $\forall x \in X \exists y \in Y, (x, y) \in W$. This is equivalent to saying the map $(x, y) \mapsto x : W \to X$ is surjective: that is, that the image of said map is all of $X$.
Now we have $\exists_{\pi_1} g = \exists_{\pi_1} (g_1 \lor \cdots \lor g_n) = (\exists_{\pi_1} g_1) \lor \cdots \lor (\exists_{\pi_1} g_1) = S_1 \lor \cdots \lor S_n = X$. To prove this, recall that $\exists_{\pi_1}$ is the left adjoint to pullback and thus commutes with joins, and that whenever $b \circ c$ is monic, we have $\exists_b c = b \circ c$.
The next step in set theory is showing $\forall x \in X \forall y_1, y_2 \in Y, (x, y_1) \in W \land (x, y_2) \in W \to y_1 = y_2$. In other words, we should show that $(x, y) \mapsto x : W \to X$ is injective.
We will show that $\pi_1 \circ g$ is mono. Suppose we had enough $h, p : K \to W$ such that $\pi_1 \circ g \circ h = \pi_1 \circ g \circ p$. Let $e : E \to K$ be the equaliser of $h, p$.
Now in $Sub(K)$, we have
$\begin{equation}
\begin{split}
K &= K \land K \\
&= h^*(W) \land p^*(W) \\
&= h^*(\bigvee_i S_i) \land p^*(\bigvee_j S_j) \\
&= \bigvee_i h^*(S_i) \land  \bigvee_j p^*(S_j) \\
&= \bigvee_i \bigvee_j h^*(S_i) \land p^*(S_j)
\end{split}
\end{equation}$
So it suffices to show that $h^*(S_i) \land p^*(S_j) \leq E$ for all $i, j$ to show that $h = p$.
To prove this, we can show that the two maps $h’, p’ : h^*(S_i) p^*(S_j)$ both factor through the subobject $S_i \land S_j$ in the same way and are therefore equal. The details would involve naming more maps than I’d like, so I’ll leave this as an exercise.
So we know that $\pi_1 \circ g : W \to X$ is a mono whose image is all of $X$; thus, it is an isomorphism. Reversing the map gives us maps $X \to W \to Y$; the composed map $f$ is such that $f|_{S_i} = f_i$ for all $i$.
For uniqueness, if we had two such maps $f, f’ : X \to Y$, we can show that each $S_i$ factors through their equaliser, so their equaliser must be $1_X$ and the maps must be equal.
