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I need help with understanding the proof of the following Lemma(68.1, page 414) in Munkres' Topology, 2nd edition. I will be grateful if someone could take me through the proof or provide a different proof to what's found in the book.


Let $G$ be a group; Let $\{G_\alpha\}$ be a family of subgroups of $G$. If $G$ is the free product of the groups $G_\alpha$, then $G$ satisfies the following condition:

Given any group $H$ and any family of homomorphisms $h_\alpha : G_\alpha \to H$, there exists a homomorphism $h:G\to H$ whose restriction to $G_\alpha$ equals $h_\alpha$, for each $\alpha$. Furthermore, $h$ is unique.

Thanks.

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    @Arturo,@dls: In my experience both patterns are reasonably common. E.g., one often proves the convergence of a recursively defined sequence by showing first what the limit must be if it exists and then using that information to prove existence. In my experience it depends on whether uniqueness is easier to show than existence (which it often is) and whether the proof of uniqueness gives insight into the proof of existence.2012-03-04

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The free product of the $G_{\alpha}$ consists of all words of the form $\mathbf{g}=g_{1\alpha_1}g_{2\alpha_2}\cdots g_{m\alpha_m}$ where $g_{i\alpha_i}\in G_{\alpha_i}$, $\alpha_i\neq\alpha_{i+1}$ for $i=1,\ldots$. Multiplication is by concatenation if the last index of $\mathbf{g}$ is different from the first index of $\mathbf{h}$; if they are equal, then we multiply $g_{m\alpha_m}$ by $h_{1\alpha_1}$, reduce as necessary, and continue.

The hardest part in proving that this is a group is proving associativity of the operation; trying to do it directly results in having to consider several different cases depending on how much cancellation there is.

Luckily, there is a better way of proving that the product is associative, known as "van der Waerden's trick." We embed this set into something that we already know is a group, and show that the product we have defined agrees with the operation on the known group.

Let $\mathbf{x}$ be a symbol that is not an element of any $G_{\alpha}$, and let $X$ consist of all words of the form $g_{1\alpha_1}g_{2\alpha_2}\cdots g_{m\alpha_m}\mathbf{x},$ with $g_{1\alpha_1}\cdots g_{m\alpha_m}$ as above. Let $S_X$ be the group of all permutations of $X$.

Let $G$ be the free product of the $G_{\alpha}$ with multiplication described as above. Given $\alpha$ and $h\in G_{\alpha}$, we define $\sigma_{h,\alpha}$ to be the element of $S_X$ given by $\sigma_{h,\alpha}(\mathbf{g}\mathbf{x}) = \left\{\begin{array}{ll} hg_{1\alpha_1}\cdots g_{m\alpha_m}\mathbf{x} &\text{if }\alpha\neq\alpha_1\text{ or }(\alpha=\alpha_1\text{ and }h\neq g_{1\alpha_1}^{-1})\\ g_{2\alpha_2}\cdots g_{m\alpha_{m}}\mathbf{x} &\text{if }\alpha=\alpha_1\text{ and }h=g_{1\alpha_1}^{-1}\\ h\mathbf{x} & \text{if }m=0. \end{array}\right.$ Given $\mathbf{h}=h_{1\beta_1}\cdots h_{m\beta_m}\in G$, we define $\sigma_{\mathbf{h}} = \sigma_{h_{1\beta_1},\beta_1}\circ\cdots\circ \sigma_{h_{m\beta_m},\beta_m}.$ (The role of $\mathbf{x}$ is as "punctuation", to ensure that there is actually an object in $X$ to which we are mapping when there is "full cancellation").

Note that the map $\mathbf{g}\mapsto\sigma_{\mathbf{g}}$ is one-to-one, since $\sigma_{\mathbf{g}} = \sigma_{\mathbf{h}}$ implies $\mathbf{g}=\sigma_{\mathbf{g}}(\mathbf{x})=\sigma_{\mathbf{h}}(\mathbf{x}) = \mathbf{h}$.

It is now easy to verify that the multiplication in $G$ satisfies $\sigma_{\mathbf{gh}} = \sigma_{\mathbf{g}}\circ\sigma_{\mathbf{h}}$; so the multiplication we have defined on $G$ is associative, since it corresponds to composition in $S_X$ and we know $S_X$ is a group. Thus, this product makes $G$ into a group.

Moreover, there is a natural embedding of each $G_{\alpha}$ into $G$, mapping $g\in G_{\alpha}$ to the word that consists only of $g$. Call this $\iota_{\alpha}\colon G_{\alpha}\to G$.

So, taking for granted that this is a group, we prove that it has the desired universal property: let $H$ be an arbitrary group, and let $h_{\alpha}\colon G_{\alpha}\to H$ be group homomorphisms from each $G_{\alpha}$. We need to show that there exists a unique homomorphism $h\colon G\to H$ such that $h_{\alpha} = h\circ \iota_{\alpha}$ for every $\alpha$.

Existence. Given $\mathbf{g}=g_{1\alpha_1}\cdots g_{m\alpha_m}\in G$ as above, we define $h(\mathbf{g}) = h_{\alpha_1}(g_{1\alpha_1})\cdots h_{\alpha_m}(g_m\alpha_m).$ This makes sense, because $g_{i\alpha_i} \in G_{\alpha_i}$ for each $i$, hence $h_{\alpha_i}(g_{i\alpha_i})$ is defined and an element of $H$; and thus the right hand side is a product of finitely many elements of $H$, hence an element of $H$.

We need to show that $h\circ\iota_{\alpha}=h_{\alpha}$ and that $h$ is a group homomorphism. The first is easy: if $g_{\alpha}\in G_{\alpha}$, then $h\circ\iota_{\alpha}(g_{\alpha}) = h(g_{\alpha}) = h_{\alpha}(g_{\alpha})$; thus, $h\circ\iota_{\alpha}$ agrees with $h_{\alpha}$ at every $g_{\alpha}\in G_{\alpha}$, hence the two are equal.

For the second, we need to show that $h$ is multiplicative. By induction, it suffices to look at products $g_{1\alpha_1}g_{2\alpha_2}$. If $\alpha_1=\alpha_2$, then we have $h(g_{1\alpha_1}g_{2\alpha_1}) = h_{\alpha_1}(g_{1\alpha_1}g_{2\alpha_1}) = h_{\alpha_1}(g_{1\alpha_1})h_{\alpha_1}(g_{2\alpha_1}) = h(g_{1\alpha_1})h(g_{2\alpha_1}),$ since $h_{\alpha_1}$ is assumed to be a group homomorphism. If $\alpha_1\neq\alpha_2$, then $h(g_{1\alpha_1}g_{2\alpha_2}) = h_{\alpha_1}(g_{1\alpha_1})h_{\alpha_2}(g_{2\alpha_2}) = h(g_{1\alpha_1})h(g_{2\alpha_2}).$ From this, one can easily show by induction on the number of letters in the product that $h$ is multiplicative. Thus, a homomorphism exists.

To prove uniqueness, suppose that $k\colon G\to H$ is a homomorphism such that $k\circ\iota_{\alpha} = h_{\alpha}$ for every $\alpha$. We need to show that $k=h$.

Since $G$ is generated by the words $g_{\alpha} = \iota_{\alpha}(g_{\alpha})$, it suffices to show that $k$ and $h$ agree on those elements. But this is now immediate: we have: $k(g_{\alpha}) = k(\iota_{\alpha}(g_{\alpha})) = h_{\alpha}(g_{\alpha}) = h(g_{\alpha}).$ Since $k$ and $h$ are both group homomorphisms and they agree on a generating set for $G$< then $k=h$. Thus, the map is unique.

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    Thanks very much for your help.2012-03-05