Why does the coproduct equal to the union in the category of subsets? Let $C$ be the category of subsets where the morphisms among the objects are inclusions. The claim here is that the coproduct in this category is simply the ordinary union. That is,
If $A$ and $B$ are objects in $C$, then:
$$ \text{Coproduct}(A,B) = A \cup B$$
Now, following the definition of the coproduct on Wikipedia,  how do I show that the morphism $f:\text{Coporduct}(A,B) \to Y$ is unique iff $\text{Coporduct}(A,B) = A \cup B$?
 A: It's helpful to think of a more "universal property" style characterisation of the union of sets in general:

The union $A\cup B$ of sets $A$ and $B$ is the smallest set containing both $A$ and $B$.

which says two things:

*

*it contains both $A$ and $B$; that is, $A\subseteq A\cup B$ and $B\subseteq A\cup B$

*it is the smallest set with this property (this is the "universal" part); that is, if for some set $U$ we have $A\subseteq U$ and $B\subseteq U$, then it follows that $A\cup B\subseteq U$
If we translate this into the language of categories by taking $C$ to be a category of (sub)sets, where the only morphisms are inclusion maps $X\to Y$ when $X\subseteq Y$, then the above two points translate to characterise $A\cup B$ as:

*

*it contains both $A$ and $B$; that is, we have morphisms $A\to A\cup B$ and $B\to A\cup B$

*it is the smallest set with this property; that is, if for some set $U$ we have morphisms $A\to U$ and $B\to U$, then there is a morphism $A\cup B\to U$.

This is almost the characterisation of the coproduct of $A$ and $B$. However, there are two (possible) issues:

*

*Does this morphism $A\cup B\to U$ come from factoring $A\to U$ and $B\to U$ through $A\cup B$?

*Is this morphism unique?

However, both of these have the same answer: in the category $C$, a morphism $X\to Y$ is unique when it exists: if $X\subseteq Y$, then the inclusion map $X\hookrightarrow Y$ is the unique morphism $X\to Y$ defined in $C$.
Therefore, the answer to (1) is yes because the two maps $A\to U$ and $A\to A\cup B\to U$ are both the unique arrow $A\to U$. Similarly, the answer to (2) is yes because if a morphism $A\cup B\to U$ exists, it is unique by definition!
