Definition of a coproduct and its universal property - connection?

Question

I have a problem connecting the definition of a coproduct with its often mentionend universal property. Let's start with the definition (just for two objects):

Let $A_1$ and $A_2$ be objects of a category $\mathcal{C}$. A coproduct of $A_1$ and $A_2$ in $\mathcal{C}$ is a triple $(A,p_1,p_2)$ where $A \in ob(\mathcal{C})$ and $p_i \in hom_\mathcal{C}(A_i,A)$ such that if $B$ is any object in $\mathcal{C}$ and $f_i \in hom_\mathcal{C}(A_i, B), i=1,2$, then there exists a unique $f \in hom_\mathcal{C}(A,B)$ such that the diagrams are commutative ($f_i = p_if$).

My apologies, I can't draw the diagram but it's pretty simple with the equation and I assume well known. While this already sounds like a universal property to me, referred to as the universal property of the coproduct is something else:

$$hom(A_1 \amalg A_2, B) \cong hom(A_1,B) \prod hom(A_2,B)$$

Now I wonder how to connect these two, e.g. how to derive this statement from the definition. It comes out of the blue for me, and I didn't came across an explanation nor a proof so far. Also often I see $\mathrm{Hom}$ instead of $hom$ - that is the same, isn't it?

Answer 1

Let $x$ and $y$ be objects of a category $C$. Their coproduct is an object $x \sqcup y$ together with morphisms $x \to x \sqcup y$ and $y \to x \sqcup y$, satisfying the universal property that given any object $z$ with maps $x \to z$ and $y \to z$, there exists a unique morphism $x \sqcup y \to z$ through which the given maps factor. Another way to say this is that the data of

• an element of the set $\mathrm{Hom}(x \sqcup y, z)$

is the same as the data of

• an element of the set $\mathrm{Hom}(x, z)$,
• and an element of the set $\mathrm{Hom}(y, z)$.

Of course, this is the same as the data of

• an element of the product of the sets $\mathrm{Hom}(x, z) \times \mathrm{Hom}(y, z)$.