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 $\{a_1,…,a_n\}$.
Best Answer
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.