Pushout of Z isomorphic to Z

Question

$\mathbb{Z}\xleftarrow{\text{id}}\mathbb{Z}\xrightarrow{\text{x2}}\mathbb{Z}$

How do we prove this group pushout is isomorphic to $\mathbb{Z}$?- I know we can take $\langle x|\rangle$ as a presentation of the identity copy of $\mathbb{Z}$,but I am unsure what the presentation of the other group is, and how to prove with the presentations that it is isomorphic to $\mathbb{Z}$

Answer 1

I will add indices to "copies" of $\mathbb{Z}$ for it to be less ambiguous. You want the pushout of $\operatorname{id}: \mathbb{Z} \to \mathbb{Z}_1$ and $2 \cdot: \mathbb{Z} \to \mathbb{Z}_2$.
Just take $\mathbb{Z}_3$ (another copy of $\mathbb{Z}$) with the arrows $2 \cdot: \mathbb{Z}_1 \to \mathbb{Z}_3$ and $\operatorname{id}: \mathbb{Z}_2 \to \mathbb{Z}_3$.

Then for any group $G$ endowed with $f_1: \mathbb{Z}_1 \to G$ and $f_2: \mathbb{Z}_2 \to G$ such that $f_1(n) = f_2(2n)$, the only factorisation through $\mathbb{Z}_3$ possible is $g=f_2$ since you need to ensure $g \circ \operatorname{id} = f_2$, and it fits since $g \circ \operatorname{id} = f_2$ and $g \circ (2 \cdot) (n)= f_2(2n) = f_1(n)$.

So you have a pushout that is $\mathbb{Z}$! And as a colimit any other pushout will be isomorphic to $\mathbb{Z}$.

Answer 2

The pushout of $A \xleftarrow{id} A \xrightarrow{f} B$ is always isomorphic to $B$, in any category. More precisely, $$\require{AMScd} \begin{CD} A @>{id}>> A \\ @V{f}VV @VV{f}V \\ B @>>{id}> B \end{CD}$$ is a pushout square, as can be verified directly using the definition.

PS: This is one of many instances where examples obfuscate what is really happening.