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?



Fundamental group of twisted loop space
**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.](https://mathoverflow.net/q/132339/8103) 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 $\{a\in \pi\_1(M) \mid f\_\*(a)=a\}$.


