Is it true that the forgetful functor $\mathcal{F}:Grp\rightarrow Set$ defined by $(A,+)\mapsto\mathcal{F}(A,+)=A$ and $A\stackrel{f}{\rightarrow}B\mapsto \mathcal{F}A\stackrel{\mathcal{F}f}{\rightarrow}\mathcal{F}B$, where $a\mapsto \mathcal{F}f(a)=f(a)$ has as universal object $(\mathbb{Z},1)$?
I have prove it as follows: given $G\in Grp$ and $g\in\mathcal{F}G$, then it exist a unique $f:\mathbb{Z}\rightarrow G$ such that $0\mapsto e$ and $1\mapsto g$. So $\mathcal{F}f(1)=f(1)=g$.
Am I missing something? Thank you.
Yes, this is correct. The unique homomorphism $f$ is just defined by $f(n)=g^n$ for $n\in\mathbb{Z}$.