# Munkres lemma 68.5

I'm reading Munkres Topology and I'm stuck in lemma 68.5 as you can see he uses the theorem 68.4 in order to imply that there is a isomorphism between $$G$$ and $$G'$$, but in order for this theorem to be applied we must have that $$\{i_{\alpha}(G_{\alpha})\}$$ and $$\{i'_{\alpha}(G_{\alpha})\}$$ generate $$G$$ and $$G'$$ which for $$G'$$ is just fine because it is stated that is a free product of the groups $$\{i'_{\alpha}(G_{\alpha})\}$$ but how do we know that $$G$$ can be generated by $$\{i_{\alpha}(G_{\alpha})\}$$? We say that the groups $$G_\alpha$$ generate $$G$$ if every element $$x$$ of $$G$$ can be written as a finite sum of elements of the groups $$G_\alpha$$.

Lemma 68.5. Let $$\{G_{\alpha}\}_{\alpha \in J}$$ be a family of groups; let $$G$$ be a group; let $$i_{\alpha} : G_{\alpha} \rightarrow G$$ 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 $$i_{\alpha}(G_{\alpha})$$.

Proof. We first show that each $$i_{\alpha}$$ is a monomorphism. Given an index $$\beta$$, let us set $$H = G_{\beta}$$. Let $$k_{\alpha} : G_{\alpha} \rightarrow H$$ be the identity if $$\alpha = \beta$$, and the trivial homomorphism if $$\alpha \neq \beta$$. Let $$h : G \rightarrow H$$ be the homomorphism given by the extension condition. Then $$h \circ i_{\beta} = h_{\beta}$$, so that $$i_{\beta}$$ is injective. By Theorem 68.2, there exists a group $$G'$$ and a family $$i'_{\alpha} : G_{\alpha} \rightarrow G'$$ of monomorphisms such that $$G'$$ is the free product of the groups $$i'_{\alpha}(G_{\alpha})$$. Both $$G$$ and $$G'$$ have the extension property of Lemma 68.3. $$\color{red}{\text{The preceding theorem then implies that there is an isomorphism}}$$ $$\phi : G \rightarrow G'$$ such that $$\phi \circ i_{\alpha} = i'_{\alpha}$$. It follows at once that $$G$$ is the free product of the groups $$i'_{\alpha}(G_{\alpha})$$.

Theorem 68.4 (Uniqueness of free products).} Let $$\{G_{\alpha}\}_{\alpha \in J}$$ be a family of groups. Suppose $$G$$ and $$G'$$ are groups and $$i_{\alpha} : G_{\alpha} \rightarrow G$$ and $$i'_{\alpha} : G_{\alpha} \rightarrow G'$$ 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 $$\phi : G \rightarrow G'$$ such that $$\phi \circ i_{\alpha} = i'_{\alpha}$$ for all $$\alpha$$.

I think it is a mistake, and it should be assumed that $$\{i_\alpha(G_\alpha)\}$$ generate $$G$$ in Lemma 68.5.

Consider situation where the indexing set $$J=\{1\}$$ is a singleton and $$G_1=\mathbb{Z}$$. Now let $$G=\mathbb{Z}\times\mathbb{Z}$$ and $$i_1:\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$$ be given by $$i_1(x)=(x,0)$$. Of course $$i_1(G_1)$$ does not generate $$G$$. So what I did here is I added $$\mathbb{Z}$$ to $$\mathbb{Z}$$, but the choice is arbitrary. We can also consider $$\mathbb{Z}\times K$$ for any group $$K$$.

Now, if $$H$$ is any group and $$h_1:\mathbb{Z}\to H$$ is a homomorphism, then we can simply define $$h:\mathbb{Z}\times\mathbb{Z}\to H$$ by $$h(x,y)=h_1(x)$$. Then $$h_1=h\circ i_1$$ and so $$G$$ together with $$i_1$$ clearly satisfies the extension property, but $$G=\mathbb{Z}\times\mathbb{Z}$$ is not (free product of) $$\mathbb{Z}$$.

Note that this reasoning can be generalized to any indexing set $$J$$. The point of this argument is that we can always add something to $$G$$ via for example direct product (which rarely gives a free product, in fact we can always choose $$K$$ to guarantee that) and retain the extension property. Without increasing the number of $$i_\alpha$$ homomorphisms. And so being generated by $$\{i_\alpha(G_\alpha)\}$$ has to be assumed.

The Lemma is correct, but Theorem 68.4 (as stated by Munkres) is not sufficient to prove it. Let us analyze Lemma 68.3. The "universal property" of the free product is

$$(*)$$ Given a group $$H$$ and a 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.

Unfortunately it is a bit unclear what the universal property is. Actually it is not only $$(*)$$, but is $$(*)$$ plus uniqueness. In other word, it is

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

This is a special case of the universal property of the coproduct (aka categorical sum) in category theory. I shall not go into details here, you can read anby book about category theory.

Note that $$(\#)$$ is a property applicable to any system $$(G,i_\alpha)$$ consisting of a group $$G$$ and a family of homomorphisms $$i_\alpha : G_\alpha \to G$$. Lemma 69.3 proves that a certain "special system" has this property.

Let us prove Theorem 68.4 in the adequate generalized form:

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 homomorphisms both satisfying $$(\#)$$. Then there is a unique isomorphism $$\phi : G \to G'$$ such that $$\phi \circ i_\alpha = i'_\alpha$$ for all $$\alpha$$.

Note that we do neither require that the $$i_\alpha, i'_\alpha$$ are monomorphisms nor that the families $$\{i_{\alpha}(G_{\alpha})\}$$ and $$\{i'_{\alpha}(G_{\alpha})\}$$ generate $$G$$ and $$G'$$, respectively.

Proof of the genearalized theorem.

By $$(\#)$$ there exist unique homomorphims $$\phi : G \to G'$$ and $$\psi : G' \to G$$ such that $$\phi \circ i_\alpha = i'_\alpha$$ and $$\psi \circ i'_\alpha = i_\alpha$$ for all $$\alpha$$. Hence $$(\psi \circ \phi) \circ i_\alpha = i_\alpha = id_G \circ i_\alpha$$ for all $$\alpha$$. Again by $$(\#)$$ (uniqueness!) we see that $$\psi \circ \phi = id_G$$. Similarly $$\phi \circ \psi = id_{G'}$$ which shows that $$\phi$$ is isomorphism.

The proof of Lemma 68.5 is now easy.

By Theorem 68.2, there exists a group $$G'$$ and a family $$i'_{\alpha} : G_{\alpha} \rightarrow G'$$ of monomorphisms such that $$G'$$ is the free product of the groups $$i'_{\alpha}(G_{\alpha})$$. By Lemma 68.3 the system $$(G',i'_{\alpha})$$ satisfies $$(\#)$$. By Theorem 68.4 there exists a unique isomorphism $$\phi : G \rightarrow G'$$ such that $$\phi \circ i_{\alpha} = i'_{\alpha}$$. It is now trivial that the $$i_{\alpha}$$ are monomorphisms.