Let $F : \text {Grp} \rightarrow \text {Set} $ be the forgetful functor. I am trying to show that $F$ is co-representable
My definition of co-representability reads:
A functor $F: C \rightarrow \text{Set}$ is called co-representable if there exists an object $Y\in C$ with $ h^Y:=\operatorname {Hom}_C(Y,-) \cong F$. How do I continue form here? I guess I need to construct an explicit isomorphism but I am not sure how
As pointed out in the comments, $\mathbb{Z}$ is the "co-representing" object. You want to prove that $Hom(\mathbb{Z}, - ) \cong F$; thus, you need to find a natural isomorphism between the two functors. This means that, for each object $G \in Grp$, you want to find a $Set$-isomorphism (i.e. a bijection) $$\eta_G : Hom(\mathbb{Z},G) \to G$$ such that for every group homomorphism $f : G \to X$ we have $$\eta_H \circ Hom(\mathbb{Z},f) = F(f) \circ \eta_G$$ As you pointed out, for each $G \in Grp$, the definition of $\eta_G$ should be $\eta_G(f) = f(1)$ for every $f \in Hom(\mathbb{Z},G)$. It's easy to see that, for every $G$, this map is a bijection (every morphism from $\mathbb{Z}$ is determined by where it sends $1$ and every $g \in G$ is a good choice). To check naturality: First note that, for each $f : G \to H$, $F(f) = f$ and $Hom(\mathbb{Z},f) = f \circ -$. Now, it's easy to check that $$\forall g \in Hom(\mathbb{Z},G) \quad (f\circ \eta_G)(g) = (\eta_H \circ (f \circ -))(g) $$