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If I am given a pointed pair of spaces $(X,A,x_{0})$ and define $P(X;x_{0},A) \subset X^{I}$ as the subspace given by the paths $\alpha$ in $X$ such that $\alpha(0) = x_{0} \text{ and } \alpha(1) \in A$. This is a pointed space $\kappa_{x_{0}}: I \rightarrow X$ as the basepoint.

Q: Prove that there are natural isomorphisms:

$$\pi_{n+1}(X,A) \cong \pi_{n}(P(X;x_{0},A),\kappa_{x_{0}}), \space \space n \geq 1$$

and conclude that $\pi_{n}(X,A)$ is a group for $n \geq 2$ and abelian if $n \geq 3$

Here I must admit I am having a difficult time interpreting what $P(X;x_{0},A) \subset X^{I}$ even means. I see that we are saying that $P(X;x_{0},A) \subset I \rightarrow X$, is this saying: that the whole of $P(X;x_{0},A) \subset I$ which is mapped to X? What is trying to be conveyed here? It looks as if there is a homotopy of some sort but I don't know what of.

I am extremely interested in this question as somehow we are going from Homotopy groups of one dimension $ \pi_{n+1}$ to that of one dimension lower $ \pi_{n}$ and I am very curious as to how this works.

Thanks much for any thoughts.


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up vote 1 down vote accepted

I won't give the full answer but a sketch of a proof of a slightly easier case of this proposition (which is a bit too much for a comment).

Let $P(X;x_0)$ just be the usual pathspace based at $x_0$. That is, $$P(X;x_0)=P(X;x_0,X)=\{ \gamma\in X^I\mid \gamma(0)=x_0\}.$$

Are you aware of the special case of your result for when $A=\{x_0\}$? In this case, $P(X;x_0,A)=\Omega(X;x_0)$, the based loop space of $X$ based at $x_0$.

The standard proof in this case is to construct a fibration $\pi \colon P(X;x_0,\{x_0\})\rightarrow X$ with fiber $\Omega(X;x_0)$ in the obvious way by mapping a path $\gamma\colon I\rightarrow X$ to its end point $\gamma(1)$. Then, consider the long exact sequence in homotopy of this fibration $$\pi_{n+1}(P(X;x_0),c_{x_0}) \stackrel{\pi_*}{\rightarrow} \pi_{n+1}(X,x_0) \rightarrow \pi_n(\Omega(X;x_0),c_{x_0}) \stackrel{\pi_*}{\rightarrow} \pi_n(P(X;x_0),c_{x_0})$$ where $c_{x_0}\colon I\rightarrow X$ is the constant path at $x_0$. You then need to prove a small lemma that $P(X;x_0,\{x_0\})$ is a contractible space and you're done.

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