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I read that there are long exact sequences of homotopy groups for each pair of pointed spaces $(X,A,x_{0})$. Now I know that for an exact sequence that, as the example below denotes $f \text{ and } g$ as maps that $im(f) = ker(g)$:

$$\pi_{n}(A,x_{0})\overset{f}\rightarrow \pi_{n}(X,x_{0})\overset{g}\rightarrow \pi_{n}(X,A)$$

now I know that $im(f)$ is all of $f$ as $A \subseteq X$ and we are just taking basepoint $x_{0}$ to itself. Thus we would have to have that $ker(g) = \pi_{n}(A,x_{0})$. I am not sure what is going on, or how to visualize this next part but for is $\pi_{n}(X,A)$ the space X with A as a basepoint? That somehow A is collapsed? I know that from the fundamental theorem of isomorphism as applied to exact sequences that $A\overset{f}\rightarrow X \overset{g}\rightarrow X/A$ but am not sure how to show that this holds with the exact sequence of homotopy groups (although they are groups). So how would I explicitly prove that this is exact? But again, what is the interpretation of $\pi_{n}(X,A)$?

So that is the first thing I am trying to understand. Next, I can somehow create a long exact sequence of this. I know what it is supposed to look like, i.e.:

$$\pi_{n}(A,x_{0})\overset{f}\rightarrow \pi_{n}(X,x_{0})\overset{g}\rightarrow \pi_{n}(X,A) \overset{h_{*}} \rightarrow \pi_{n-1}(A,x_{0})\overset{f}\rightarrow \pi_{n-1}(X,x_{0})\overset{g}\rightarrow \pi_{n-1}(X,A) .... \pi_{0}(A,x_{0})\overset{f}\rightarrow \pi_{0}(X,x_{0})\overset{g}\rightarrow \pi_{0}(X,A)$$

I however don't understand its meaning - especially as it is moving from groups of different dimensions such as some loop space in n dimensions to that of n-1 dimensions etc.

Your insight is much appreciated.

Thanks,

Brian

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2 Answers 2

up vote 1 down vote accepted

There is a correction needed to your question: $f(\pi_n(A,x_0))$ is not the same as $\pi_n(A,x_0)$ because for the latter, homotopies of maps are taken in $A$ whereas in the former they are taken in $X$.

I feel that to understand this type of exact sequence it helps to study a simpler and algebraic example, namely that of the exact sequence of a fibration of groupoids, which models the bottom end of the exact sequence you give, and yields more precise information there. This notion of fibration is developed in Section 7.2 of Topology and Groupoids. A morphism $p: E \to B$ of groupoids is a fibration if for each $x \in Ob(E)$ and each $b$ in $B$ starting at $y= p(x)$ there is an $ e \in E$ starting at $x$ such that $p(e)=b$. (The original definition with applications was in the paper R. Brown, J. Algebra 15 (1970) 103-132.) The exact sequence is at 7.2.9, and is of the form

$$ F_x(x) \to E(x) \to B(y) \to \pi_0 F_x \to \pi_0 E \to \pi_0 B, $$

where $F_x= p^{-1}(y)$ is the fibre of $p$ over $y$. In this situation $F_x(x) \to E(x)$ is injective so it does not quite model the general situation.

This notion reduces to a covering morphism of groupoids if in the above definition of fibration the lifting $e$ is unique, which is equivalent to each $F_x(x)$ is the trivial group.

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Thanks Ronnie - I am guessing your the "R. Brown in the paper mentioned"? Now I see that $f(\pi_n(A,x_0))$ is a mapping to $\pi_n(A,x_0)$. Because of the mapping we are in X and not A, hence we are dealing with the homotopies in X. I am a bit lost on your definition of fibration as I understood a fiber of a point $b \in B$ is the subspace of E which maps to $b$. I don't understand how $E$ and $B$ relate to $X$ and $A$, but am guessing that somehow we are making a group structure on $X$ and $A$ which are $B$ and $E$ respectively. Is this correct? Thanks again. –  Relative0 Aug 18 '13 at 20:02

To respond to your last comment: "I however don't understand its meaning - especially as it is moving from groups of different dimensions such as some loop space in n dimensions to that of n-1 dimensions etc" perhaps it would be helpful to consider an example. Consider a disk $D$ with boundary $\partial D = S^1$. There is a nonzero class in $\pi_2(D,S^1)$ (I am omitting the basepoint to simplify notation), represented by the identity mapping of the disk. The boundary of this is the class in $\pi_1(S^1)$ represented by the identity mapping of the boundary circle. So the idea of moving to a different (lower) dimension is actually quite natural.

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