I am working on algebraic topology. I am trying to prove the Wirtinger presentation using the Van Kampen theorem.
However, I have some difficulties understanding the concept of free product with amalgamation.
I read that :
Let $G_1 = \langle R_{G_1} \mid S_{G_1} \rangle$ and $G_2 = \langle R_{G_2} \mid S_{G_2} \rangle$ be presentations of $G_1$ et $G_2$. Assume we have $\Phi : G \rightarrow G_1$ et $\Psi : G \rightarrow G_2$ injective homomorphisms.
\Then a presentation of the free product of $G_1$ with $G_2$ over $G$ is : $$\langle R_{G_1} \cup R_{G_2} \mid S_{G_1} \cup S_{G_2} \cup [\forall x \in G, \Phi(x) = \Psi(x)] \rangle.$$
But what happens when we have no such injective homomorphism?
Here is my problem : what is $\langle x_1, …, x_n \rangle \star_{\langle x_1, …, x_n \mid R_+, R_- \rangle} \{1\}$ equal to ?
I assume (intuitively) that it is $\langle x_1, …, x_n \mid R_+, R_- \rangle$, but how to prove it, since $\langle x_1, …, x_n \mid R_+, R_- \rangle$ afford no injective homomorphism in $\{1\}$ ?
Thank you very much.
AF
Best Answer
The amalgamated free product is a pushout in the category of groups.
Given (any) two group maps $f : G \to G_1, g : G \to G_2$, the amalgamated product $G_1 *_{G} G_2$ is a group with arrows $G_1 \to G_1 *_{G} G_2$ and $G_2 \to G_1 *_{G} G_2$ such that this triple is universal with respect to the property that
$\require{AMScd}$ \begin{CD} G_1 *_G G_2 @<<< G_2\\ @AAA @AAA\\ G_1 @<<< G \end{CD} is commutative.
It can be described internally as :
$G_1 *_G G_2 = G_1 * G_2/<f(a)g(a)^{-1} : a \in G>$ with the arrows from $G_1$ and $G_2$ being the canonical "embeddings".
In your question, there is no canonical map from $<a_1,..,a_n | R_+, R_->$ to $<a_1,..,a_n>$. The former is a quotient of the latter. You must specify the map first before forming the product.