Why is $\ 1*_{\mathbb Z}\mathbb Z =\mathbb Z/2\mathbb Z=\{±1\}$?

Question

Why is $\ 1*_{\mathbb Z}\mathbb Z =\mathbb Z/2\mathbb Z=\{±1\}$ ?

Our Professor got this result as the Fundamental group of $\mathbb RP^2$ by using Van-Kampen Theorem

$*_{\mathbb Z}$ is the amalgamation

$\textbf{My Question}:$ the proof is clear, but the amalgamation not.

I think you can give me a short explanation, otherwise i can also write the whole proof.

edit: $\textbf{THE PROOF}$:

we parametrized $S^2$

$S^1\times[-\pi/2,\pi/2]\rightarrow S^2$

$(x,t)\mapsto cost(x,0)+sint(0,0,1)$

Let $U_1=p(upper hemisphere)$$\quad$p:projection from $S^2$ to $\mathbb RP^2$

$U_2=p(S^1\times (-\pi/4,\pi/4))$ then,

$U_1\cap U_2\simeq$ Cylinder

Then we used Van-Kampen

$\underbrace{\pi_1(U_1\cap U_2)}_{\mathbb Z}\longrightarrow\underbrace{\pi_1(U_1)}_{=1}$

$\downarrow$$\qquad$$\qquad$$\qquad$$\qquad$$\downarrow$

$\underbrace{\pi_1(U_2)}_{=\mathbb Z}\longrightarrow\pi_1(\mathbb RP^2)$

That's all then he wrote the result.

$\textbf{Here is the necessary part which i only need}$

According to the definition

If you have two mappings, $\psi_1,\psi_2$ with

$\psi_1:\mathbb Z\rightarrow 1$

$\psi_2:\mathbb Z\rightarrow\mathbb Z$

then $1*_{\mathbb Z}\mathbb Z=(1*\mathbb Z)/N$ where N is the normal subgroup generated by the elements of the form $\psi_1(a)\psi_2(a)\in 1*\mathbb Z$

and this must be equal to $\mathbb Z/2\mathbb Z$, and i ask WHY

Answer 1

Issues with the proof aside, here is the answer to your question. Let $A = 1$, $B = \langle b \rangle \cong \Bbb Z$, $A \cap B = \langle c \rangle \cong \Bbb Z$.

Let $\psi_1$, $\psi_2$ be the inclusion maps from $A \cap B$ to $A$, $B$ respectively. Since $A$ is trivial, $\psi_1$ is trivial too. As loops, the generator of $A \cap B$ goes twice around the generator of $B$. Hence $\psi_2(c) = b^2$.

We have $$ A * B = \langle b \rangle, $$ and $$ N = \langle b^2 \rangle. $$

Thus $$ (A * B) / N = \langle b \mid b^2 \rangle \cong \Bbb Z / 2 \Bbb Z $$

Answer 2

In general $A *_C B$ means: you are given homomorphisms $\alpha: C \to A$ and $\beta:C \to B$ (these are not typically included in the notation, but are a crucial part of the data!). You then take a the free product of $A$ and $B$, and finally, take the maximal quotient in which $\alpha(c)$ and $\beta(c)$ are identified, for all $c \in C$.

Now if $A = 1$, the trivial group, then this is just the same as taking the maximal quotient of $B$ in which $\beta(C)$ is identified to $1$; equivalently, the quotient of $B$ by the normal subgroup generated by $\beta(C)$.

Now if $B$ and $C$ are $\mathbb Z$ (which is in particular abelian, so that any subgroup is normal), and you are told that $1 *_C B = \mathbb Z/2 \mathbb Z$, then the image of $\beta$ must be $2\mathbb Z$. This pretty much pins down $\beta$ (it fixes it up to a sign).

This should let you reverse engineer the problem/proof you are trying to understand.