# On the converse of extension lemma of external free product --- Munkres Lemma 68.5

This is Lemma 68.5 in Munkres, which states as follows

Let $$\{G_{\alpha}\}_{\alpha\in J}$$ be a family of groups; let $$G$$ be a group; let $$i_{\alpha}:G_{\alpha}\longrightarrow G$$ be a family of homomorphisms. If the below extension condition $$(*)$$ holds, then each $$i_{\alpha}$$ is an injective homomorphism and $$G$$ is the free product of the groups $$i_{\alpha}(G_{\alpha})$$.

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

To prove this, we need the following uniqueness theorem:

Let $$\{G_{\alpha}\}_{\alpha\in J}$$ be a family of groups. Suppose $$G$$ and $$G'$$ are groups and $$i_{\alpha}:G_{\alpha}\longrightarrow G$$ and $$i_{\alpha}':G_{\alpha}\longrightarrow G$$ are families of injective homomorphisms, such that the families $$\{i_{\alpha}(G_{\alpha})\}$$ and $$\{i_{\alpha}'(G_{\alpha})\}$$ generate $$G$$ and $$G'$$, respectively. Now if both $$G$$ and $$G'$$ have the extension property stated in the extension theorem, then there is a unique isomorphism $$\phi:G\longrightarrow G'$$ such that $$\phi\circ i_{\alpha}=i_{\alpha}'$$ for all $$\alpha$$.

Munkres firstly prove that $$i_{\alpha}$$ is an injective homomorphism for each $$\alpha$$, and I have no problem with it. Then he tried to show that $$G$$ is the free product of the groups $$i_{\alpha}(G_{\alpha})$$, as follows:

There exists a group $$G'$$ and a family $$i_{\alpha}':G_{\alpha}\longrightarrow G'$$ of injective homomorphisms such that $$G'$$ is the free product of the groups $$i'_{\alpha}(G_{\alpha})$$. Thus it satisfies the extension theorem and thus the uniqueness theorem implies that there is an isomorphism $$\phi:G\longrightarrow G'$$ such that $$\phi\circ i_{\alpha}=i'_{\alpha}$$. It then follows that $$G$$ is the free product of the groups $$i_{\alpha}(G_{\alpha})$$.

I don't quite understand the conclusion here, by his argument, we have $$G\cong G'=\prod_{\alpha\in J}^{*}i'_{\alpha}(G_{\alpha}),$$ so at most we have $$G=\prod_{\alpha\in J}^{*}\phi\circ i_{\alpha}(G_{\alpha}),$$ how did he get from here to the conclusion that $$G$$ is the free product of $$i_{\alpha}(G)$$?

Thank you!

Edit 1: Proof of isomorphism preserving free product structure:

As Lee Mosher pointed out, we need to check if the isomorphism preserving free product structure, to pass the $$\phi^{-1}$$ from outside the free product to the inside.

After some thought, I generated a possible way to show isomorphism preserving free product. Also, as Lee Mosher said, we do need the isomorphism property, although homomorphism is good enough to preserve the operation. We need a map to be isomorphism to preserve the reduced word property and the $$1-1$$ correspond between the image and preimage, perhaps.

To show the isomorphism preserves the free product structure, we need another way to realize the free product of groups. The definition we have used until now follows from Munkres, and he claimed that using this definition will make life much easier in the proof of the existence of free products, since verifying group properties, especially the associativity, will be really tedious and irritating if we use the definition I am going to talk about.

However, to show the isomorphism preserving group property, we need to specify the group property and group elements, in which case we need following alternative but equivalent definition of free product.

[Alternative Definition] Let $$W$$ denote the set of all reduced words in the elements of the groups $$G_{\alpha}$$. Then think of $$G$$ as being simply the set $$W$$ itself, with the product of two words simply obtained by juxtaposing them and reducing the result. So the group operation is simply the product of elements. The identity element $$1$$ corresponds to the empty word, and each group $$G_{\beta}$$ corresponds to the subset of $$W$$ consisting of the empty set and all words of length $$1$$ of the form $$(x)$$ for $$x\in G_{\beta}$$ and $$x\neq 1_{\beta}$$.

Now we proceed to the proof:

Then for every $$x\in G'$$, it can be written as $$x=x_{1}\cdots x_{n}$$ where $$x_{i}\in i_{\alpha_{i}}'(G_{\alpha_{i}})$$ and $$\alpha_{i}\neq \alpha_{i+1}$$ for all $$i$$. Then since $$\phi^{-1}$$ is a homomorphism, we have $$\phi^{-1}(x)=\phi^{-1}(x_{1}\cdots x_{n})=\phi^{-1}(x_{1})\cdots\phi^{-1}(x_{n})$$. Now note that by the property of $$\phi^{-1}$$: $$i_{\alpha}=\phi^{-1}\circ i_{\alpha}'$$, it sends each $$x_{i}\in i'_{\alpha_{i}}(G_{\alpha_{i}})$$ back to $$i_{\alpha_{i}}(G_{\alpha_{i}})$$, so $$\phi^{-1}(x)$$ is an element which can be presented by words in $$\phi^{-1}\circ i_{\alpha_{i}}'(G_{\alpha_{i}})$$, namely $$(\phi^{-1}(x_{1}),\cdots,\phi^{-1}(x_{n}))$$, where $$\phi^{-1}(x_{i})\in i_{\alpha_{i}}(G_{\alpha_{i}})$$ for each $$i$$, and $$\alpha_{i}\neq \alpha_{i+1}$$ for each $$i$$. Those properties are preserved since $$\phi$$ is an isomorphism.

The above shows that $$\phi^{-1}\Big(\prod_{\alpha\in J}^{*}i_{\alpha}'(G_{\alpha})\Big)\subset \prod_{\alpha\in J}^{*}\phi^{-1}\circ i_{\alpha}'(G_{\alpha}).$$

Conversely, for every $$y$$ in the free product of the RHS of the above inclusion, it can be written as $$y=y_{1}y_{2}\cdots y_{m}$$ where $$y_{i}\in \phi^{-1}\circ i_{\alpha_{i}}'(G_{\alpha_{i}})$$ and $$\alpha_{i}\neq \alpha_{i+1}$$ for all $$i$$. Since $$\phi^{-1}$$ is an isomorphism, for each $$y_{i}$$, there is a unique $$x_{i}\in i_{\alpha_{i}}'(G_{\alpha_{i}})$$ such that $$y_{i}=\phi^{-1}(x_{i})$$.

Thus, $$y$$ can be also written as $$y=\phi^{-1}(x_{1})\phi^{-1}(x_{2})\cdots\phi^{-1}(x_{m})$$, and since $$\phi^{-1}$$ is a homomorphism, we have $$y=\phi^{-1}(x_{1}x_{2}\cdots x_{m})$$ where $$x_{i}\in i_{\alpha_{i}}'(G_{\alpha_{i}})$$ and $$\alpha_{i}\neq\alpha_{i+1}$$.

This shows the inverse inclusion and we conclude the proof.

Note that I am really not sure if my proof is correct since it is a little bit confusing here.. so please do not hesitate to point out any mistake and to correct me if you want :) Thank you!

This equation in your question is incorrect: $$G=\prod_{\alpha\in J}^{*}\phi\circ i_{\alpha}(G_{\alpha})$$ Instead, starting from the equation $$G'=\prod_{\alpha\in J}^{*}i'_{\alpha}(G_{\alpha})$$ you should first substitute using $$i'_\alpha = \phi \circ i_\alpha$$ to obtain $$G'=\prod_{\alpha\in J}^{*}\phi \circ i_\alpha(G_{\alpha})$$ and then you should apply the inverse isomorphism $$\phi^{-1} : G' \to G$$ on both sides to obtain $$G=\prod_{\alpha\in J}^{*}\phi^{-1} \circ \phi \circ i_\alpha(G_\alpha) = \prod_{\alpha\in J}^{*} i_\alpha(G_\alpha)$$
• Oh I believe one note is worth to make here, we can pass $\phi^{-1}$ from the outside of the free product to the inside of the free product is because $\phi^{-1}$ is a homomorphism. Commented Nov 13, 2019 at 19:39
• I think I would say that you need more than just that $\phi^{-1}$ is a homomorphism, you need that it is an isomorphism. But you're right, there's something to be checked here, along the lines of the slogan that "isomorphisms preserve free product structure". Commented Nov 13, 2019 at 20:00