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Feb 27, 2021 at 14:48 vote accept Luigi M
Feb 26, 2021 at 8:41 answer added Mark Grant timeline score: 4
Feb 26, 2021 at 1:02 comment added Will Sawin In this case we want a map from $[0,1]$ to a space of maps from $[0,1]$ to $M$, so we want a map from the square to $M$, and we know what the map should do on three of the sides, so we can just fill in by retracting the square to the union of those three sides. Sorry I meant $f(\sigma)$, the action of the automorphism $f$ on the loop $\sigma$.
Feb 26, 2021 at 1:01 comment added Will Sawin The action is of $\pi_1(M, x_0)$ on $\pi_0 (\Omega_f(M))$ - probably a typo. The rule for this action is the same as any fibration - you take a map from $[0,1]$ to the base $M$ sending $0$ and $1$ to $x_0$, you lift the $0$ to a point in any fixed connected component of the fiber, then you lift the map from $[0,1]$, and you see what component $1$ ends up in.
Feb 26, 2021 at 0:57 comment added Luigi M @WillSawin I think I'm having an hard time visualising the action of $\pi_1(M,x_0)$ on $\pi_1(\Omega_f(M))$. As you suggested I'm thinking of a loop in $\Omega_f(M)$ as an homotopy between some loops $\alpha$ and $f(\alpha)$. How am I suppose to see the action of $\pi_1(M)$? and what's $\gamma(\sigma)$ that you mentioned?
Feb 25, 2021 at 22:48 comment added Luigi M Thanks a lot! I will need some time to digest these comments though.
Feb 25, 2021 at 22:45 comment added Will Sawin $\pi_0(\Omega_f(M))$ is the set of orbits in $\pi_1(M)$ of the $\gamma$-conjugacy action, i.e. $g_1$ and $g_2$ are in the same orbit if there exists $h$ with $ \gamma(h) g_1 h^{-1} =g_2$. The image of the map $\pi_1( \Omega_f(M), x_0) \to \pi_1(M, x_0)$ consists of only the $\gamma$-invariant elements of $\pi_1 (M , x_0)$ - it's not so hard to see how a homotopy between a loop representing an element of $\pi_1(M)$ and its image under $\Gamma$ gives a loop in $\Omega_f(M)$.
Feb 25, 2021 at 22:40 comment added Will Sawin Think about what happens when you take a loop in $\Sigma_f(M)$ starting at $x_0$ and ending at $x_0$ and flow it along a loop $\sigma$ in $M$. We can flow by simply moving the starting point along the map of $\sigma$ and the ending point along the path of $\gamma (\sigma)$, and extending the path to follow the new starting and ending points. The final result is our original path, composed with $\gamma(\sigma)$ on the left and $\sigma^{-1}$ on the right. This shows how $\pi_1$ acts on $\pi_0$, and gives several consequences:
Feb 25, 2021 at 22:35 comment added Luigi M @WillSawin I'm definitely not sure, I was thinking in terms of connected components but I might be -very- wrong
Feb 25, 2021 at 22:32 comment added Will Sawin Are you sure the last map is an injection? I would think $\pi_1(M, x_0 ) \to \pi_0 (\Omega_{x_0} M)$ is something like $\sigma \to f(\sigma) \sigma^{-1}$, which is not the zero map.
Feb 25, 2021 at 21:57 history asked Luigi M CC BY-SA 4.0