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A loop in a topological space $X$ based at $a\in X$ is a continuous function $\rho: [0,1]\rightarrow X$ such that $\rho(0)=\rho(1)=a$.

The reverse of $\rho$ is $\overline{\rho}$ such that $\overline{\rho}(s)=\rho(1-s)$ for each $s\in [0,1]$. Clearly $\rho$ is also a loop based at $a$.

It seems rather obvious that $\rho$ is homotopic to $\rho$, i.e. there exists a continuous function $H:[0,1]\times [0,1]\rightarrow X$ such that $H(0,t)=H(1,t)=a$, $H(s,0)=\rho(s)$, and $H(s,1)=\overline{\rho}(s)$, but I'm not sure how to define/construct such a homotopy?

Sid Caroline
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    It is not true in general. If this occurs, then $\rho^2$ is the identity in the fundamental group, so that it must have an element of order 2. This doesn't always happen. – Randall Oct 06 '17 at 02:25
  • @Randall That should be an answer –  Oct 06 '17 at 02:26
  • An example where this is true, is the projective plane. – jMdA Apr 08 '21 at 10:49

1 Answers1

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If $\rho \simeq \overline{\rho}$ then $\rho^2$ is the identity in $\pi_1(X)$. You are asking this to hold for all loops $\rho$. This is a lot to ask and is usually not so. Of course it is true trivially for simply connected spaces, but otherwise you are demanding that all non-identity elements in $\pi_1(X)$ have order 2. This is not so and is far from true. For a concrete example, $\pi_1(S^1) \cong \mathbb{Z}$ has no elements of order 2.

Randall
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