Fundamental group of twisted loop space

Question

I'm interested in computing the fundamental group of the twisted loop space $$\Omega_f(M)=\{ \gamma \in C^{\infty}(\Bbb R,M) \mid \gamma(s+1)=f\gamma(s)\}$$ where $f \in \text{Aut}(M,x_0)$, for example a diffeomorphism with a fixed point $x_0$.

The twisted loop space is part of a fibration $$\Omega_{x_0}M \to \Omega_f(M) \xrightarrow{ev_0} M$$ where the map $ev_0$ is the evaluation map at $s=0$ and $\Omega_{x_0}M$ is the loop space (based at $x_0$).

The relevant piece of the l.e.s. homotopy gives $$\pi_2(M,x_0)\xrightarrow{\partial} \pi_1(\Omega_{x_0}M,x_0)\to \pi_1(\Omega_f(M),x_0) \to \pi_1(M,x_0)\to \pi_0(\Omega_{x_0}M)\to \pi_0(\Omega_f(M))$$

The last map should be an injection, hence we have a surjection $\pi_1(\Omega_f(M),x_0) \twoheadrightarrow \pi_1(M,x_0)$. I don't quite understand the effect of the boundary map from the second homotopy group of $M$, so I cannot really go on. As far as I know, if $M$ is simply connected then $\pi_1(\Omega_f(M),x_0) \cong \pi_2(M,x_0)$. If $\delta=0$ that would be the case but I cannot verify it.

Is this computation done somewhere so I can check how's done?

Answer 1

Edit: The following is incorrect, see below. It might be more useful to think of $\Omega_f(M)$ as sitting in a homotopy pullback $\require{AMScd}$ \begin{CD} \Omega_f(M) @>>> M\\ @V V V @VVfV\\ M @>>\operatorname{Id}> M \end{CD}

Then you can make use of the "Mayer-Vietoris sequence" of homotopy groups $$\cdots \to \pi_2(M, x_0) \stackrel{\partial}{\to} \pi_1(\Omega_f(M),x_0) \to \pi_1(M, x_0) \times \pi_1(M, x_0) \to \pi_1(M,x_0) \to \cdots$$ described in the answers here. The last map above is $(a,b)\mapsto a\cdot f_*(b)^{-1}$, which is not a homomorphism in general but is a surjection of pointed sets. This immmediately shows for example that when $\pi_2(M,x_0)=0$ then $\pi_1(\Omega_f(M),x_0)$ is isomorphic to the graph of the homomorphism $f_*:\pi_1(M,x_0)\to \pi_1(M,x_0)$.

To go further you'd have to analyse the connecting map $\partial: \pi_2(M,x_0)\to \pi_1(\Omega_f(M),x_0)$. It may be (though I'm not sure) that the map preceding it is given by $(x,y)\mapsto x-f_*(y)$, in which case $\partial=0$. A good place to read up on this sequence seems to be May and Ponto's More concise algebraic topology, Section 2.2

Edit: as Tyrone points out, I wrote down the wrong pullback square, and consequently my conclusions were incorrect. What I should have written down was the topological pullback

$\require{AMScd}$ \begin{CD} \Omega_f(M) @>>> C^\infty(\mathbb{R},M)\\ @V V V @VV\operatorname{ev}_{0,1}V\\ M @>>(\operatorname{Id},f)> M\times M \end{CD} in which the map $\operatorname{ev}_{0,1}$ is homotopically equivalent to the diagonal map $\triangle: M\to M\times M$. Then the MV sequence looks like

$$\cdots \to \pi_2(M \times M) \stackrel{\partial}{\to} \pi_1(\Omega_f(M)) \to \pi_1(M \times M) \to \pi_1(M\times M) \to \cdots$$

where this time the last map is given by $(a,b)\mapsto(a b^{-1}, f_*(a) b^{-1})$. Then in the case where $\pi_2(M)=0$, for example, we can see that $\pi_1(\Omega_f(M))$ is isomorphic to the subgroup $\{a\in \pi_1(M) \mid f_*(a)=a\}$.