Compare the universal mapping property with the extension lemma of free groups --- Munkres Lemma 69.1

Question

I am following Munkres book to study free group, but he did things pretty different than other online note or book.

When talked about free group, he gave the extension lemma:

Let $G$ be a group; let $\{a_{\alpha}\}_{\alpha\in J}$ be a family of elements of $G$. If $G$ is a free group with system of free generators $\{a_{\alpha}\}$, then $G$ satisfies the following condition:

$(*)$ Given any group $H$ and any family $\{y_{\alpha}\}$ of elements of $H$, there is a homomorphism $h:G\longrightarrow H$ such that $h(a_{\alpha})=y_{\alpha}$ for each $\alpha$.

Furthermore, $h$ is unique. Conversely, if the extension condition $(*)$ holds, then $G$ is a free group with system of free generators $\{a_{\alpha}\}$.

However, I also saw the universal mapping property of free group from some online notes, which is following:

Let $G$ be a group with a generating set $X\subset G$. Then $G$ is free on $X$ if and only if given any group $H$ and map $\phi:X\longrightarrow H$, $\phi$ can be extended to a unique homomorphism $\phi^{*}:G\longrightarrow H$ so that $\phi=\phi^{*}\circ j$ where $j:X\longrightarrow G$ is the inclusion map.

My question is:

(1) Is the extension lemma equivalent to the universal mapping property?

(2) In the extension lemma, the index of the family of elements of $H$ is the same as the index of the free generators, what does this mean? In other word, what does the homomorphism do if $\{y_{\alpha}\}$ has the number of element less than the number of free generator?

Thank you!

Answer 1

1. I think when he says "and any family $\{y_\alpha\}$,", he means to imply that again $\alpha \in J$, i.e., that the number of elements chosen in $H$ is the same as the number chosen in $G$ (and is indexed by the same set); otherwise the equality at the end of the starred line doesn't make sense.

1A. To see that (with this assumption) the two are equivalent: let $\phi$ be a map as in the second definition, and let $y_\alpha = \phi(x_\alpha)$ for each $\alpha \in J$. Then you have the setup in the first definition, which guarantees you a unique homomorphism $h$ matching $\phi$ on the set of $x$s. This map $h$ serves as $\phi^{*}$ in the second definition.

For the other direction, suppose you have the set of $y$s as in the first definition. Then for $\alpha \in J$, define $\phi(x_\alpha) = y_\alpha$. This gives you a map from $X = \{x_\alpha\}_{\alpha \in J}$ to $H$. The map $\phi^{*}$ guaranteed by the second definition is then the map $h$ required in the first.