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By Seifert-Van Kampen theorem, $\pi_1(P) = \mathbb Z/2$. If $a$ is a generator of the group, this means that $aa = 1$. In terms of homotopy, this implies that $c_{x_0} \cong w*w$, where $w \in a$ is a loop. How can I visually construct homotopy between these two paths? i.e. make a square-like diagram, where the $x$-axis represents the space $P$ and $y$-axis represents time interval.

James C
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  • Do the answers to https://math.stackexchange.com/questions/629595/fundamental-group-of-projective-plane-is-c-2?rq=1 help? – Alex J Best Mar 15 '21 at 02:34
  • Which definition of $P$ do you use? – Paul Frost Mar 15 '21 at 15:43
  • @paul The picture looks like: draw a circle, and points are identified antipodally. The disk with quotient topology is P. So, the edges along the hemisphere represent $a$. I haven’t had time to look at Best’s link above yet. – James C Mar 15 '21 at 21:20

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Let $q : D^2 \to P$ be the quotient map which identifies antipodal points of $S^1$. We can regard $P$ as the adjunction space $D^2 \cup_\phi S^1$, where $\phi : S^1 \to S^1, \phi(z) = z^2$.

Let $u : [0,1] \to D^2, u(t) = e^{\pi it}$ and $v : [0,1] \to D^2, v(t) = -e^{\pi it}$. Then $q \circ u = q \circ v$. Let us denote this map by $w : [0,1] \to P$. It is a loop based at $q(1)$ representing the generator of $\pi_1(P,q(1))$. The path $\gamma = u * v$ is a loop in $D^2$ (it runs once counterclockwise around $S^1 \subset D^2$) based at $1 \in S^1$ and $$H: [0,1] \times [0,1] \to D^2, H(t,s) = (1-s)\gamma(t)+ s$$ is a path homotopy to the constant path based at $1$. Consider the homotopy $H' = q \circ H : [0,1] \times [0,1] \to P$. This is a homotopy of paths in $P$. We have $H'(t,0) = q \circ (u * v) = (q \circ u) * (q \circ u) = w * w$ and $H'(t,1) = q(1)$.

This proves your claim.

Paul Frost
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  • Thanks for the answer, @Paul. One thing that I haven’t understood is, is the homotopy well-defined? First, $H$ is a homotopy on $D^2$ that is equivalent to a straight path between $\gamma(t)$ and $1 \in S^1$. For fixed $t_0 \in [0, \frac{1}{2}]$, the $H(t_0, s)$ and $H(t_0 + \frac{1}{2}, s)$ are two distinct paths. Going back to $P$, since antipodal points on the circle are identified together, $\gamma(t) \sim \gamma(t+\frac{1}{2})$. Thus, $H’(t_0, 0) = H’(t_0+\frac{1}{2}, 0)$. This implies that $H’(t_0, s) = H’(t_0 + \frac{1}{2}, s)$ for every $s \in I$. However, they are different paths! – James C Mar 16 '21 at 00:04
  • I mean, if $s \in (0, 1)$, then I already know that $H(t_0, s) \neq H(t_0 + \frac{1}{2}, s)$ (they lie on interior of the disk, and they are different points.) Thus, $q \circ H(t_0, s) = H(t_0, s) \neq H(t_0 + \frac{1}{2}, s) = q \circ H(t_0 + \frac{1}{2}, s)$, contradiction. – James C Mar 16 '21 at 00:07
  • Given $\gamma$, the only thing we have to know is $H(0,s) = H(1,s)$ for all $s$. This is true because $\gamma(0) = \gamma(1)$. It means that $H$ is a homotopy of paths in $D^2$. And whatever $H$ does, we can project it via $q$ to $P$. – Paul Frost Mar 16 '21 at 00:13
  • Yes, I was ignoring a basic fact that $q \circ H$ is a composition of continuous map, so it defines a homotopy on $P$. Nevertheless, I am confused about the counterintuitive behavior that I described. I try to visualize homotopy by tracing a path that a particle moves along. The particle here is $\gamma(t_0) = \gamma(t_0 + \frac{1}{2})$. However, one particle cannot move along two distinct paths. Is such phenomenon happening because $\gamma$ is not an injective path? – James C Mar 16 '21 at 00:32
  • $\gamma$ is in fact not injective, but the only two points of $[0,1]$ having the same image under $\gamma$ are $0$ and $1$. But the point is that for any two closed paths $\gamma, \delta : [0,1] \to D^2$ based at the same $x \in D^2$ we get $\gamma \simeq \delta$ by $H(t,s) = (1-s)\gamma(t) + s \delta(t)$, simply because $D^2$ is convex. – Paul Frost Mar 16 '21 at 00:40
  • Just a final comment, $q \circ \gamma$ is injective, right? $q \circ \gamma(t_0) = q \circ \gamma(t_0 + \frac{1}{2})$. (To be fair, I said $\gamma$ is injective earlier, not that $q \circ \gamma$ is injective.) – James C Mar 16 '21 at 08:07
  • No, $q \circ \gamma$ is not injective: As you say, $q \circ \gamma(t_0) = q \circ \gamma(t_0 +1/2)$. This is obviuous because we want $q \circ \gamma = w * w$. And, by the way, no closed path can be inijective since it maps $0, 1$ to the same point. But this is a "trivial" non-injectivity which is due to closedness. – Paul Frost Mar 16 '21 at 09:16
  • oh god. I am have been thinking “non-injectivity” all the time and typed out the word “injective” instead. I have no idea what I have been thinking. Thank you for the answer. – James C Mar 16 '21 at 09:21