How to show a diagram is a pushout? Let the $U$, $V$ be two subobjects of an object $A$ of a topos. I want to show the diagram
$$\require{AMScd}
\begin{CD}
U\cap V @>>>U \\
@VVV @VVV \\
V@>>>U\cup V
\end{CD}$$
is a push out.
I tried use the universal propery to show if there is a $X$ with two arrows one from $U$ to $X$ and the other from $V$ to $X$ then there shold be a unique arrow from $U\cup V$ to $X$....
I know a pushout is a colimit and any topos, contains all finite colimits.
 A: Suppose we have morphisms $f : U \to X$ and $g : V \to X$ such that $f |_{U\cap V} = g |_{U \cap V}$.  Now consider the subobjects $\Gamma_f, \Gamma_g$ of $A \times X$ given by $(\operatorname{inc}, f) : U \to A \times X$ and $(\operatorname{inc}, g) : V \to A \times X$ respectively (the graphs of $f$ and $g$), and then the subobject $\Gamma_f \cup \Gamma_g$ of $A \times X$.
Lemma: The morphism $\pi_1 : \Gamma_f \cup \Gamma_g \to A$ is a monomorphism.
Proof: Suppose we have a test object $T$ and two morphisms $h_1, h_2 : T \to \Gamma_f \cup \Gamma_g$ such that $\pi_1 \circ h_1 = \pi_1 \circ h_2$.  Then $h_1^{-1}(\Gamma_f) \cup h_1^{-1}(\Gamma_g) = h_2^{-1}(\Gamma_f) \cup h_2^{-1}(\Gamma_g) = T$.  So, by the distributivity of the subobject lattice of $T$, we see that
$$(h_1^{-1}(\Gamma_f) \cap h_2^{-1}(\Gamma_f)) \cup \\
(h_1^{-1}(\Gamma_f) \cap h_2^{-1}(\Gamma_g)) \cup \\
(h_1^{-1}(\Gamma_g) \cap h_2^{-1}(\Gamma_f)) \cup \\
(h_1^{-1}(\Gamma_g) \cap h_2^{-1}(\Gamma_g)) = T.$$
Now, since $\pi_1 : \Gamma_f \to A$ is a monomorphism (in fact, it is isomorphic to $U$), we conclude that $h_1 = h_2$ when restricted to $h_1^{-1}(\Gamma_f) \cap h_2^{-1}(\Gamma_f)$.  Similarly, $h_1 = h_2$ when restricted to $h_1^{-1}(\Gamma_g) \cap h_2^{-1}(\Gamma_g)$.  Also, on $h_1^{-1}(\Gamma_f) \cap h_2^{-1}(\Gamma_g)$, $\pi_1 \circ h_1$ factors through $U$ and the equal map $\pi_1 \circ h_2$ factors through $V$, so $\pi_1 \circ h_1 = \pi_1 \circ h_2$ factors through $U \cap V$; as a consequence, we see that $h_1$ and $h_2$ both factor through the graph of $f |_{U\cap V} = g |_{U\cap V}$, implying that $h_1 = h_2$ on this set also.  Similarly, $h_1 = h_2$ on $h_1^{-1}(\Gamma_g) \cap h_2^{-1}(\Gamma_f)$.  In conclusion, we see that $h_1 = h_2$ on all of $T$ (more formally, the equalizer of $h_1$ and $h_2$ contains all four subsets of $T$, so it is equal to all of $T$).  $~\square$
From here, we see that the image of $\pi_1 : \Gamma_f \cup \Gamma_g\to A$ must be equal to $\pi_1(\Gamma_f) \cup \pi_1(\Gamma_g) = U \cup V$, so in fact $\Gamma_f \cup \Gamma_g$ is isomorphic to $U \cup V$.  We can now form the composition of $\pi_2$ with the inverse of this isomorphism to get the desired morphism $U\cap V \to X$ extending both $f$ and $g$.
The uniqueness of the extension map to $U \cup V$, on the other hand, is easy: if we have two extensions, then their equalizer contains both $U$ and $V$, so the equalizer also contains $U \cup V$.
(Note that the proof of the lemma above is a straightforward rendering of the proof resulting from the formal proof in the internal language of a topos dividing into four cases for whether two points $(a, x), (a', x') : A \times X$  which are in $\Gamma_f \cup \Gamma_g$ are in $\Gamma_f$ (and thus $a\in U \land x = f(a)$ respectively $a'\in U \land x' = f(a')$) or in $\Gamma_g$ (and thus $a\in V \land x = g(a)$ respectively $a'\in V \land x' = g(a')$).  And note that $\pi_1((a, x)) = \pi_1((a', x'))$ reduces to $a = a'$.)
