Universal Property...The Map is not Well-defined?

Question

How does the following proposition from this book make sense:

Proposition 2.8 Let $G$ be the group defined by the presentation $(X,R)$. For any group $H$ and map of sets $\alpha : X \to H$ sending each element of $R$ to $1$, there exists a unique homomorphism $G \to H$ making the following diagram commute:

\begin{array}{rcl} X &\rightarrow & G \\ & \searrow & \downarrow \\ & & H \end{array}

Sorry: I don't know how to write commutative diagrams in latex...see page 36 of the linked document. How could $\alpha$ map every element of $R$ to the identity when it is possible that $R$ is not contained in $X$? I want to use this proposition show that $A$ with the presentation $\langle a_1,...,a_n \mid [a_i,a_j ] \rangle$ is isomorphic to the free abelian group on $n$ elements $b_1,..,b_n$, but this one point is confusing me at the moment. The most natural thing to do is to let $\alpha(a_i) = b_i$. But such a definition doesn't obviously satisfy $\alpha (R) = \{1\}$, since $\{[a_i,a_j] \mid i,j =1,...,n\}$ is not a subset of $\{a_1,...,a_n\}$.

Answer 1

The author writes "sending each element of $R$ to $1$ (in the obvious sense)" and adds the following footnote:

Each element of $R$ represents an element of $FX$, and the condition requires that the unique extension of $\alpha$ to $FX$ sends each of these elements to $1$.

This should answer your question.

Answer 2

The text should have been clearer on this point, but to make sense of this, consider each element of $R$ consists of a word in elements of $X$ and their inverses, and extend $\alpha$ to all such words by multiplication in the group $H$. For example, if $x,y \in X$ and we consider the word $x^2 y^3 x^{-4} y^{-3}$, then $\alpha(x^2 y^3 x^{-4} y^{-3}) = \alpha(x)^2 \alpha(y)^3 \alpha(x)^{-4} \alpha(y)^{-3}$.