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First: here are a couple links of which I am looking at. I try to add the relevant information (at least to my understanding) from them.

http://topospaces.subwiki.org/wiki/Actions_of_the_fundamental_group

http://mathoverflow.net/questions/19775/different-way-to-view-action-of-fundamental-group-on-higher-homotopy-groups

This second link goes in to some amount of how to view the action, but a "different way" and I am still trying to understand "any way" to see this.

I know that we can consider the iterated loop space $\Omega^{k}(X,x_{0})$. This is obtained by iterating the loop space construction k times where at each stage, the new basepoint is chosen as the $\textit{ constant }$ loop taking the old basepoint as its value.

I don't understand how that works and taking the "constant loop" to be the old basepoint as its value. What I imagine is that as with the loop space of $[0,1]/1\sim 0 \text{ which is } I^{1}/\partial I \cong S^{1}$ is just a point? I could be way off here, but could use some help on what the constant loop really is. This is just the first (side question), but it comes up in the fundamental groups action on higher homotopy groups.

I could use some help in understanding the action of $\pi_{1}(X,x_{0})$ on $\pi_{n}(X,x_{0})$. Now I know that $\pi_{1}$ is an n-dimensional group which consists of $\pi_{1}$, so if I were to collapse $\pi_{1}$ to a point it is going to reduce $\pi_{n} \text{ to } \pi_{n-1}$. Hence this is the reason that we get the result on the bottom of this page: http://topospaces.subwiki.org/wiki/Loop_space_of_a_based_topological_space - that $\pi_{k}(X,x_{0}) \cong \pi_{k-1}(\Omega(X,x_{0}))$ But what does this do?

In the end I see that somehow one can show (not homework to turn in) that: "for homotopy classes $\alpha, \beta, \in \pi_{1}(X, x_{0} ) \text{ and } \omega \in \pi_{n}(X, x_{0} )$ we have: $$\alpha · (\beta · \omega) = (\alpha · \beta) · \omega$$

Intuitively I can imagine that the $\pi_{1} \text{ or } S^{1}$ part of the loop space that I can get to any point on $S^{n-1}$ no matter where I start from on $S^{1}$. I don't know if this is correct though, but somehow seems it should be as we can get to $\Omega^{k}(X,x_{0})$ by looking at the underlying loop space (and that they are all connected).

I apologize if I made a lot of mistakes in my thoughts but I want to give a background on my thought process.

Thanks,

Brian

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Anyone have some deeper insight to this question? –  Relative0 Aug 18 '13 at 14:14
1  
What does "$\pi_1$ is an $n$-dimensional group which consists of $\pi_1$" mean? –  Leon Lampret Apr 25 at 19:56

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