# Clarification on Lemma 68.5 of Munkres

In James R. Munkres's "Topology" he has the lemma

"Lemma 68.5. Let $\{G_{\alpha}\}_{\alpha \in J}$ be a family of groups; let $G$ be a group; let $i_{\alpha}: G_{\alpha} \to G$ be a family of homomorphisms. If the extension condition of Lemma $68.3$ holds, then each $i_{\alpha}$ is a monomorphism and $G$ is the free product of the groups $i_{\alpha}(G_{\alpha})$."

"Lemma 68.3. Let $\{G_{\alpha}\}_{\alpha \in J}$ be a family of groups; let $G$ be a group; let $i_{\alpha}: G_{\alpha} \to G$ be a family of homomorphisms. If each $i_{\alpha}$ is a monomorphism and $G$ is the free product of the groups $i_{\alpha}(G_{\alpha})$, then $G$ satisfies the following condition:

Given a group $H$ and family of homomorphisms $h_{\alpha}: G_{\alpha} \to H$, there exists a homomorphism $h: G\to H$ such that $h \circ i_{\alpha}=h_{\alpha}$ for each $\alpha$.

Furthermore h is unique."

The proof of Lemma 68.5. uses the following lemma

"Lemma 68.4. Let $\{G_{\alpha}\}_{\alpha \in J}$ be a family of groups. Suppose $G$ and $G'$ are groups and $i_{\alpha}: G_{\alpha} \to G$ and $i_{\alpha}': G_{\alpha} \to G'$ are families of monomorphisms such that the families $\{i_{\alpha}(G_{\alpha})\}$ and $\{i_{\alpha}(G_{\alpha}')\}$ generate $G$ and $G'$, respectively. If both $G$ and $G'$ have the extension property stated in the preceding lemma, then there is a unique isomorphism $\phi: G\to G'$ such that $\phi \circ i_{\alpha} = i_{\alpha}'$ for all $\alpha$."

My question is whether the statement of Lemma 68.5. is missing the additional assumption that the family $\{i_{\alpha}(G_{\alpha})\}$ generates $G$? Or is that implicitly implied by the condition in Lemma 68.3.?

• No, the extension property is not enough (just think of trivial groups $G_\alpha$ as an extreme example); he just forgot the generation condition. Commented Sep 26, 2017 at 3:02
• @MoisheCohen: If $G$ has the extension property for the trivial groups $G_\alpha$, then $G$ has to be trivial. Otherwise take $H:=G$, then $h$ is not unique.
– j.p.
Commented May 23, 2018 at 5:56
• The extension property is enough. If $G$ is the free product and $H$ has the extension property, then one can show (using the extension property in both directions) that $G$ can be seen as subgroup of $H$ and $H$ is the semi-direct product between the kernel of the map from $H$ to $G$ and $G$ as subgroup of $H$.
– j.p.
Commented May 23, 2018 at 6:00

It comes down to what the extension property means. If it includes the uniqueness of $$h$$, then the extension property itself is enough. Otherwise it's not.
Based on the phrasing of Theorem 68.4, it is reasonable to assume that extension property does not include the uniqueness of $$h$$.