Showing $\Bbb{Z}*\Bbb{Z}\cong F_2$ using universal properties

Question

I am just a beginner in Algebraic Topology, and I need to show $\Bbb{Z}*\Bbb{Z}\cong F_2$, where $F_2$ is the free group on two generators, using universal properties.

Using simple constructions, I see both $F_2$ and $\Bbb{Z}*\Bbb{Z}$ look like the collection of all words of the form $\{a^{\alpha_1}\cdot b^{\beta_1}\cdot a^{\alpha_2}\cdot b^{\beta_2}\dots a^{\alpha_n}\cdot b^{\beta_n}|\alpha_i,\beta_i\in\Bbb{Z}\}$.

But I need to prove by universal properties, so first I will list the relevant ones:

The universal property of $\Bbb{Z}*\Bbb{Z}$:

There are two inclusions $i:\Bbb{Z}\rightarrow\Bbb{Z}*\Bbb{Z}$ and $j:\Bbb{Z}\rightarrow\Bbb{Z}*\Bbb{Z}$, and for any group $G$ with two homomorphisms $\phi:\Bbb{Z}\rightarrow G$ and $\psi:\Bbb{Z}\rightarrow G$ there is a unique homomorphism $\eta:\Bbb{Z}*\Bbb{Z}\rightarrow G$ such that $\eta\circ i=\phi$ and $\eta\circ j=\psi$.

The universal property of $F_2$:

Given a set of two elements $S=\{a,b\}$, $F_2$ is the free group on $S$, if it has a mapping $i:S\rightarrow F_2$ and for any group $G$ with a mapping $\phi:S\rightarrow G$ there is a unique homomorphism $\psi:F_2\rightarrow G$ such that $\psi\circ i=\phi$.

Now, I have no idea how to proceed. How are proofs by universal properties go about?

Answer 1

You can try to show that $\mathbb{Z}*\mathbb{Z}$ satisfies the universal property of $F_2$.

$\mathbb{Z}*\mathbb{Z}$ satisfies the universal property of $F_2$.

Let $S=\{a,b\}$ define $i(a)=1_{\mathbb{Z}_1}$ and $i(b)=1_{\mathbb{Z}_2}$, for any morphism $f:S\rightarrow G$, you can define $f_1:\mathbb{Z}\rightarrow G$ such that $f_1(n)=f(a)^n$ and $f_2(n)=f(b)^n$, the universal properties of $\mathbb{Z}*\mathbb{Z}$ implies the existence of a morphism of group $h:\mathbb{Z}*\mathbb{Z}\rightarrow G$ such that $h(n*0)=f(a)^n$ and $h(0*m)=f(b)^m$, in particular $h(i(a))=f(a)$ and $h(i(b))=f(b)$.

This implies that $F_2$ and $\mathbb{Z}*\mathbb{Z}$ are isomorphic.

Answer 2

The universal property of free groups is really an explicit version of the statement that the free group functor $F: \mathbf{Set} \to \mathbf{Grp}$ is left-adjoint to the forgetful functor $U: \mathbf{Grp} \to \mathbf{Set}$.

Now the coproduct in the category of sets is the disjoint union, whereas the coproduct in the category of groups is the free product. Using the fact that left-adjoint functors commute with colimits, we have: $F(\{a,b\}) = F(\{a\} \coprod \{b\}) \cong F(\{a\}) * F(\{a\}) \cong \mathbb Z * \mathbb Z$. Here we used the obvious fact that $\mathbb Z$ is the free group on one generator.

This proof may too high-level for you right now, depending on your background, but it has the nice property that it generalizes easily to sets of any size and other categories with free objects.