Universal properties for groups arising from representations Consider the forgetful functor $U:\textbf{Grp}\to\textbf{Set}$ and its left adjoint $F$.
The functor $X=\textbf{Set}(A,U(-))$ is represented by $F(A)$, i.e., $X\simeq H^{F(A)}$. By the Yoneda lemma, this gives a unique element $u\in X(F(A))$ (in fact $u=\eta_A$, the unit of the adjunction from the beginning) such that the following universal property holds: for any $p\in X(B)$ there exists a unique $p':F(A)\to B$ such that $p=U(p')\circ \eta_A$.
But the functor $U$ itself is also representable, and I'm a little confused by the universal property that arises in the same manner as above if we write out the details. Here's what I'm getting: $U$ is represented by $F(1)=\mathbb Z$ ($1$ is the singleton set), i.e., there's a natural isomorphism $U\simeq H^{F(1)}=H^\mathbb Z$. This corresponds to some $u\in U(F(1))=U(\mathbb Z)$ that satisfies this universal property:
For any arrow $p:1\to U(B)$, there is a unique arrow $p': \mathbb Z
\to B$ such that $p=U(p')\circ \eta_1$. Is this right? This looks like a castrated version of the universal property above, and I don't know why would anyone use this instead of the comprehensive universal property above. Apparently (if I understand correctly), Leinster deduces from this property that $u$ in this case should be $\pm 1$. Does this make sense (just from what I wrote, without reading Leinster)?

$U$ is representable iff $\hom(1,U(-))$ is representable, since these functors are isomorphic. So the second universal property is really a special case of the first. The first is the universal property of any free group, the second is the universal property of the free group in one generator (i.e. $\mathbb{Z}$ under addition). The universal element here is an element $u \in \mathbb{Z}$ such that $\hom(\mathbb{Z},G) \to U(G)$, $f \mapsto f(u)$ is bijective for all groups $G$, and this only works for $u=\pm 1$.
