I'm in the category $Set$:
A pullback of two insertion maps $f,g$ with the same codomain is the intersection of the two domains.
A pushout of two insertion maps $f,g$ with the same domain (which must be an intersection of the two domains, which imply they are a subset to some other bigger set) is the union of the two codomains.
But I learned that when you create dual notions, you have to flip arrows, and turn monomorphisms to epimorphisms and vice versa.
So I wanted to take a look of the dual notion of an intersection:
$$ f: Z \to X \\ g: Z \to Y $$
So if $Z$ stay's a normal set, and $f,g$ are surjections, than $X,Y$ are isomorphic to quotient sets. (I think?) pushouts are always a specific subset of the disjoint union. I almost think it is always just the disjoint union.
Is this correct? Is my assumption that union is not really the dual of intersection correct? What actually is the pushout (and pullback) of surjections?
EDIT:
Okay, so I'm making an educated guess at the dual of intersection in the category of groups.
So the coproduct in $Grp$ is the free product, and so the construct is a subgroup of the free product.
In my example I take the common domain for the product to be $C_6$, the cyclic group of order 6, and I take two normal subgroups of $C_6$, $N = \{0,3\}$, and $N' = \{0,2,4\}$ $$ f: C_6 \to \{\{0,3\},\{1,4\},\{2,5\}\} \\ g: C_6 \to \{\{0,2,4\},\{1,3,5\}\} $$
So the free group with amalgamation contains $\{\{0+N\},\{1+N\},\{2+N\},\{1+N'\},\{2+N'\},...?\}$
I'm afraid I'm not mathematically versed enough to solve this.
In the category of sets, epimorphisms are surjective functions, and they indeed correspond to quotient maps.
Namely, if $f:A\to B$ is surjective, take the equivalence relation $a\sim a'$ iff $f(a)=f(a')\ $ (called the kernel of $f$), then $f$ factors through a natural bijection $A/{}_\sim \to B$.
The pushout of general functions $f:Z\to X,\ g:Z\to Y$ is not a subset but a quotient of the disjoint union $X\sqcup Y$, namely it is 'glued' along $Z\ $ ($f(z)$ will be the same as $g(z)$ for all $z\in Z$).
Pushout of quotient maps correspond to union of equivalence relations, more precisely, the equivalence relation generated by their union.
In your example of groups, we have to take the (normal subgroup generated by the) union of the given normal subgroups, which in your case gives back the whole group, $C_6$, so the quotient will be the trivial group.
To get more convinced, pushout of a surjective homomorphism is surjective, and as both $C_3$ and $C_2$ are simple groups, we only have these surjections, up to isomorphism: $C_2\to C_2,\ C_2\to \{0\},\ C_3\to C_3,\ C_3\to \{0\}$.
Dually to the injective case, pullback of quotient maps correspond to intersection of the corresponding equivalence relations / normal subroups.