What does "coproduct of $\Bbb{Z}*\Bbb{Z}$ of $\Bbb{Z}$ by itself" mean?

Question

Prove that the group $F(\{x,y\})$ is a coproduct of $\Bbb{Z}*\Bbb{Z}$ of $\Bbb{Z}$ by itself in the category Grp.

What does "coproduct of $\Bbb{Z}*\Bbb{Z}$ of $\Bbb{Z}$ by itself" mean?

Answer 1

You know what $A * B$ is for groups $A,B$? Now put $A=B=\mathbb{Z}$. Then you know what $\mathbb{Z} * \mathbb{Z}$ is.

If $S,T$ are sets, then $F(S \sqcup T) = F(S) * F(T)$. This is formal (left adjoints preserve colimits). If $S = \{x\}$, then $F(S) = \mathbb{Z}$. It follows that $F(\{x,y\})=\mathbb{Z} * \mathbb{Z}$.

Answer 2

A way to prove that some object is a coproduct is to show that it satisfies the universal property.

The homomorphisms $\varphi$ from $\Bbb Z\ \left(\cong F(\{x\})\right)$ to any group $G$ are determined by the element $\varphi(1)\ \in G$, which can be arbitrary.

The homomorphisms $\varphi$ from $F(\{x,y\})$ to $G$ are determined by the elements $\varphi(x)$ and $\varphi(y)\ \in G$, which can be arbitrary, independently to each other.

Hence, whenever we have a pair of group morphisms $\alpha,\beta:\Bbb Z\to G$, there will be a unique map $F(\{x,y\}\to G$ [the one which picks $\alpha(1)$ for $x$ and $\beta(1)$ for $y$] that makes the coproduct diagram commutative.