I came across the following problem while studying for a topology exam: Let $M$ be a topological space, let $\Lambda(M)=M^{S^1}$, the space of continuous maps $S^1\to M$ with the compact-open topology. If $s_0\in S^1$ and $x_0\in M$, let $\Omega(M,x_0)=\{\omega\in \Lambda(M)|\omega(s_0)=x_0\}$. Assuming that $M$ is path-connected and simply-connected, show that $\pi_n(\Lambda(M),*)\simeq\pi_n(\Omega(M,x_0),*)\oplus\pi_n(M,x_0)$ for $n\geq 1$ where $*=$ the constant map at $x_0$.

$\Lambda(M)$ is the space of (unpointed) maps $S^1\to M$ and $\Omega(M,x_0)$ is the space of pointed maps $S^1\to M$, which is the loopspace of $M$. I know that $\pi_n(M,x_0)\simeq\pi_{n-1}(\Omega(M),*)$, so I am trying to prove that $\pi_n(\Lambda(M),*)\simeq \pi_{n+1}(M,x_0)\oplus\pi_n(M,x_0)$. I also know that the quotient of $\Omega(M)$ by homotopy classes of loops is $\pi_1(M)=0$, because $M$ is simply connected, so all maps in $\Omega(M)$ are nullhomotopic.

However, I don't know how to relate the homotopy groups of $M$ to $\Lambda(M)$. Any insight would be appreciated!


I think that $\Omega M \to \Lambda M \to M$ is a fiber sequence, where $\Lambda M \to M$ is the evalutation at $s_0$. Then, you have the long exact sequence of homotopy groups associated to a fibration, i.e (I will omit basepoint for sake of conciseness): $\ldots \pi_n (\Omega M) \to \pi_n(\Lambda M) \to \pi_n(M) \to\pi_{n-1}(\Omega M) \ldots$ and it should split (I am not really sure, but for example $\pi_n(\Lambda M) \to \pi_n(M)$ admits a right inverse, it remains to check the injectivity of $\pi_n (\Omega M) \to \pi_n(\Lambda M)$).

  • $\begingroup$ There is a map $M\rightarrow \Lambda M$ which is the inclusion of the constant loops $\endgroup$
    – Thomas Rot
    May 11 '16 at 19:55
  • $\begingroup$ Yeah, that's the right inverse I mention! $\endgroup$ May 11 '16 at 23:44

This paper on the arxiv looks at "Crossed modules and the homotopy 2-type of a free loop space". The monoidal closed category of crossed complexes allows for a convenient algebraic model of this 2-type.

  • 1
    $\begingroup$ If I am allowed to ask: in your opinion is my approach correct or are there any flaws? $\endgroup$ Jul 11 '14 at 12:46

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