Let $A$, $B$ be two subobjects of $X$. The intersection subobject $A \cap B$ is the pullback of $A \to X$, $B \to X$. Then we can define the union object $A \cup_1 B$ as the pushout of $A \cap B \to A$ and $A \cap B \to B$. Then there is a canonical mapping $A \cup_1 B \to X$. Also we can define the union subobject $A \cup_2 B \to X$ as the image of the canonical morphism $A \oplus B \to X$. I want to know whether $A \cup_1 B \simeq A \cup_2 B$.
Firstly $A \oplus B \to X$ has a factorization $A \oplus B \to A \cup_1 B \to X$ where the left arrow is an epimorhpism, because $A \cap B \rightrightarrows A \oplus B \to A \cup_1 B$ is a coequalizer diagram. Thus it suffices to prove that $A \cup_1 B \to X$ is a monomorphism. What I've found is just $A \to A \cup_1 B$, $B \to A \cup_1 B$, $A \cap B \to A \oplus B$ are monomorphisms.
Alternatively, we can try to prove that $A \cup_2 B$ has the universal property of pushout characterized by coequalizer diagram. I think this can be done by the Freyd-Mitchell embedding theorem, since we are only dealing with finite limits. Then $A \cup_2 B$ is the submodule $A + B$, and the canonical morphism $A \oplus B \to A + B$ has kernel $\{(x, -x) \mid x \in A \cap B\}$, which shows that $A \cap B \to A \oplus B \to A + B$ is exact, i.e. $A \cap B \rightrightarrows A \oplus B \to A + B$ is a coequalizer diagram.
Is this correct?