Could someone help me to verify the claim that the Milnor join $*$ construction https://ncatlab.org/nlab/show/Milnor+construction to define the topological space from a finite group will give the following results?
1) $$\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z} =S^1$$
2) $$\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z} =S^2$$
Here the finite group $\mathbb{Z}/2\mathbb{Z}$ of order 2 and $S^n$ is a $n$-sphere.
Do I have the correct intuition / answers above? How do we show my guesses rigorously step-by-step? Start from the very basic to the claim above.
To understand $E_n \mathbb Z/2$ as a space, note that $\mathbb Z/2 \cong S^0$.
The space $S^0 * X$ is also called the suspension of $X$, denoted by $SX$ and should be visualized as two cones on $X$ (in the link you can see the suspension of $S^1$.)
It is fairly easy to convince yourself that $SS^n \cong S^{n+1}$, and this can be proved using stereographic projection. Therefore, the $n$-th step in Milnor's construction is $E_n \mathbb Z/2=S^{n+1}$, and thus the colimit is $E \mathbb Z/2=S^\infty$.
(Sanity check: the infinite sphere is indeed contractible, as $EG$ should be)
To understand the action, remember that if $X$ is a $G$-space, then the induced action on $G*X = I\times G\times X/(0,a,y)\sim(0,b,y),(1,a,y)\sim(1,a,z)$ is given by $g.[t,h,x]=[t,gh,gx]$.
In the case where $G=\mathbb Z/2$ this means that the action of $1$ acts on the $X$ part separately and switches from one cone to the other.
It follows by simple induction that the action of $1$ on $E_n \mathbb Z/2=S^{n+1}$ is the antipodal map, $x \mapsto -x$, and therefore the same is true for the colimit $E \mathbb Z/2=S^\infty$.
(Sanity check: the antipodal map is continous and has no fixed points, thus the $\mathbb Z/2$ action is indeed continous and free)
Since $0$ acts on $E \mathbb Z/2$ trivially, we conclude that $$B \mathbb Z/2 \cong (E \mathbb Z/2)/(\mathbb Z/2) \cong S^\infty/x \sim -x \cong {\mathbb RP}^\infty$$ i.e. the infinite real projective space.