Why is $\ 1*_{\mathbb Z}\mathbb Z =\mathbb Z/2\mathbb Z=\{±1\}$? 
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
 A: 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
$$
(A * B) / N = \langle b \mid b^2 \rangle \cong \Bbb Z / 2 \Bbb Z
$$
A: 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. 
