Question about the map $S^1\to S^1$ in the context of the real projective plane $\mathbb{R}P^2$ I was recently working on an exercise to compute $H^*(\mathbb{R}P^2\times \mathbb{R}P^2,\mathbb{Z})$ and there was a particular step in the solution for which i would like to get a better intuition.
The solution says

Viewing $\mathbb{R}P^2$ as $D^2/{\sim}$ where we identify $p \in S^1 =
 \partial D^2$ with $−p$, we set $U$ to be a small disk around $0 \in  
 D^2$ and $V = \mathbb{R}P^2 − \{0\}$. Thus $U$ is contractible and $V$
deformation retracts onto the subspace $S^1/{\sim}$, which is
homeomorphic to $S^1$ through a loop joining $p$ and $−p$. The intersection $U\cap V$ is an annulus, so the diagram
$\require{AMScd}$ \begin{CD} U\cap V @>{}>> V\\ @AAA @AAA\\ S^1
 @>{f}>> S^1 \end{CD}
in which the lower map $f\colon S^1\to S^1$ is $z\mapsto z^2$,
commutes up to homotopy.

I would love to understand why the lower map needs to be chosen as $z\to z^2$.
My thoughts:
Clearly the annulus $U\cap V$ is homotopy equivalent to $S^1$ and $V$ is homotopy equivalent to $S^1/{\sim}$ which apparently seems to be homeomorphic to $S^1$ as pictured in the solution. The homotopy equivalence of $U\cap V$ and $V$ is also comprehensible by construction.
So up to homotopy i understand the top horizontal arrow and the two vertical arrows. If i imagine the bottom horizontal map $f$ to be the attaching map of the open disk onto the boundary circle $S^1$ of $\mathbb{R}P^2$ i understand why it's $z\mapsto z^2$.
Why can't we, for example, just consider the identity map $\operatorname{id}\colon S^1\to S^1$ as the bottom horizontal map. Why is the attaching map the obvious choice?
And related to that question: Is that the reason why we have chosen the disk $D^2/{\sim}$ identified with antipodal points rather than the unit circle $S^1/{\sim}$?
 A: View $S^1$ as the unit sphere in $\mathbb{C}$ and the disc $D^2$ as the cone on $S^1$. Thus the disc has coordinates $(z,t)$ where $z\in S^1$ and $t\in[0,1]$, and $S^1\times\{0\}$ is identified to a point. Then $\mathbb{R}P^2$ is the quotient of the disc formed by the identification $(z,1)\sim(-z,1)$. I will write the coordinates in $\mathbb{R}P^2=D^2/\sim$ with angular brackets.
We have
$$U=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t\leq 1/2\},\qquad V=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t>0\}.$$
The contraction of $U$ is given by $F_s\langle z,t\rangle=\langle z,(1-s)t\rangle$. The inclusion $S^1\hookrightarrow U\cap V$ is given by $z\mapsto \langle z,\frac{1}{2}\rangle$ and is a homotopy equivalence. The retraction of $V$ onto $\mathbb{R}P^1\cong \{\langle z,1\rangle\in\mathbb{R}P^2\}$ is the projection $\langle z,t\rangle\mapsto\langle z,1\rangle=\langle-z,1\rangle$, which is well-defined since $\langle z,t\rangle\in V\Rightarrow t>0$. The homotopy $G_s\langle z,t\rangle=\langle z,(1-s)t+s\rangle$ is that which is required to show that the map is a deformation retraction.
We identify $\mathbb{R}P^1$ with $S^1$ by the inverse homeomorphisms
$$\langle z,1\rangle\mapsto z^2,\qquad z\mapsto \langle \sqrt z,1\rangle=\langle-\sqrt z,1\rangle.$$
Call $\alpha: V\rightarrow S^1$ the homotopy equivalence which is the composite of the previous projection followed by this identification. Then clearly the composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha}S^1$ which appears in your diagram is the degree $2$ map
$$z\mapsto z^2.$$
Homefully you agree with this. The point of course is that we never had a chance to consider any other map to put in your square diagram: the map $f$ was what was given to use by the geometry. One thing we should note is that we have made special use of the fact that $\alpha$ is a homotopy equivalence.
If we're feeling brave we can try to see what the maps above look like in homogeneous coordinates. Take $\mathbb{R}P^2$ as a quotient of $S^2$ by the relation $(x,y,z)\simeq(-x,-y,-z)$ and write its coordinates with square brackets as $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)\in\mathbb{R}^3$ satisfy $x^2+y^2+z^2=1$.
The quotient map $\Phi:D^2\rightarrow\mathbb{R}P^2$ is given by
$$\Phi(x,y,t)=[t\cdot x,t\cdot y,\sqrt{1-t^2}],$$
and it's easily seen that this induces a homeomorphism of $D^2/\sim$ as above onto $\mathbb{R}P^2=S^2/\simeq$. The sets which appear are
$$U=\{[x,y,z]\in\mathbb{R}P^2\mid |z|\geq\sqrt{3/4}\},\qquad V=\{[x,y,z]\in\mathbb{R}P^2\mid |z|<1\}$$
(Note the change in signs of the last coordinate!). The inclusion $S^1\hookrightarrow U\cap V$ is the map $(x,y)\mapsto\Phi(x,y,1/2)=[\frac{1}{2}x,\frac{1}{2}y,\sqrt{3/4}]$. Note that this is indeed an inclusion since any coset in its image has a unique representative with positive last coordinate. The projection $V\rightarrow \mathbb{R}P^1$ is the map $[x,y,z]\mapsto[x/\sqrt{x^2+y^2},y/\sqrt{x^2+y^2},0]$ (you might like to down the homotopy $G_s$ from before using these coordinates to show again that this map is a deformation retraction). The identification $\mathbb{R}P^1\cong S^1$ is induced using stereographic projecion. In the direction we need it is
$$[x,y,0]\mapsto (x^2-y^2,2xy).$$
The composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha} S^1$ is now
$$(x,y)\mapsto(x^2-y^2,2xy).$$
(Notice that this is exactly the same map as before: $x+iy\mapsto (x+iy)^2$). I'll leave you to check any details.
So, to answer your second question: this is why it's better for this question to consider $D^2/\sim$ - everything is much easier!
