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In the last paragraph in page 345 of Hatcher's Algebraic Topology(link:http://pi.math.cornell.edu/~hatcher/AT/ATch4.pdf), Hatcher says that $\pi_1(A,x_0)$ acts on the long exact sequence of homotopy groups for $(X,A,x_0)$, the action commuting with the various maps in the sequence.

I can't see the commutativity.

For $[f] \in \pi_n(X,x_0)$, the action is defined by $[\gamma][f]=[\gamma f]$ where $\gamma f$ is the map as in the following figure (on the left), while for $[f] \in \pi_n(X,A,x_0)$, $\gamma f$ is defined as in the right figure. Also, these two are not homotopic as maps $(I^n,\partial I^n,J^{n-1})\to (X,A,x_0)$ in gerenal. Then how can the action commute with the map $\pi_n(X,x_0) \to \pi_n(X,A,x_0)$?

enter image description here

blancket
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2 Answers2

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I think all actions of $\pi_1(A,x_0)$ are the relative ones.

In particular, I think Hatcher sees the long exact sequence of $(X,A,x_0)$ as follows :

$\cdots\to\pi_n(A,x_0,x_0)\to\pi_n(X,x_0,x_0)\to\pi_n(X,A,x_0)\to\pi_{n-1}(A,x_0,x_0)\to\cdots$

This seems to be consistent with the proof of Theorem 4.3. Thus it makes sense to interpret all actions of $\pi_1(A,x_0)$ to any of the groups above by the right figure in your post.

The commutativity of the action of $\pi_1(A,x_0)$ should be clear if we stick with the relative version.

Ronnie Brown
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Aliceyuenyuen
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  • Thanks. Do you also think that the commutativity is nontrivial if the action is not given by relative ones? – blancket Jan 20 '20 at 11:49
  • I think it is not even commutative in general, if you use the absolute action on $\pi_n(X)$, and relative action on $\pi_n(X,A)$. I guess something like $X=S^2\vee S^1$, $A=x_0=p$(the connecting point), $f:S^2\to S^2, f=id$, with $\gamma$ parametrizing $S^1$, could be a possible candidate for counter example. However I have no idea how to prove this. – Aliceyuenyuen Jan 21 '20 at 06:54
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This question has another another approach to an answer which is taken in Section 6.3 of the book Nonabeian Algebraic Topology {NAT}. The basic idea is to form for a triple $(X,A,C)$ of spaces, i.e. $C \subseteq A \subseteq X$, and $C$ is thought of as a set of base points, a functor $\rho(X,A,C)$ which which in dimension $0$ is $C$; in dimension $1$ is $\pi_1(A,C)$, the fundamental groupoid of $A$ on a set of base points; and in dimension $2$ is the set of homotopy classes rel vertices of maps $$(I^2, \partial I^2, \partial \partial I^2) \to (X,A.C), $$where $\partial I^2$ is the boundary of the square $I^2$ and $\partial \partial I^2$ is the set of vertices of the square. You will notice that this structure does not form a group!

However it turns out that for each $c \in C$ the obvious map obvious map $\pi_2(X,A,c) \to \rho_2(X,A,C)$ is injective (Proposition 6.3.8). Further $\rho_2(X,A,C)$ has two compositions $+_1, +_2$ which make the whole structure into a double groupoid . There is also an extra structure of so-called connections which make it equivalent to the crossed module $$\partial: \pi_2(X,A,C)\to \pi_1(A,C).$$

Having got all this done Chapter 6 of NAT goes on to prove a 2-d Seifert Van Kampen Theorem for $\rho$ which by translation to the equivalent crossed module allows new calculations of 2nd relative homotopy groups, as pushouts of crossed modules.

The point is that the additional structures allow, once you have mastered them, clearer and more powerful proofs. For more discussion of the history and methodology, see this paper.

Ronnie Brown
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  • It is interesting that the opposition to the ideas in my answer goes back a long time (1974) and is related to issues in homotopy theory which go back further to the opposition to E.Cech's seminar on higher homotopy groups at the 1932 Zurich ICM. This conceptual dissonance is put in context in my article arxiv/abs/2001.00534, – Ronnie Brown Jan 16 '20 at 15:13
  • Not a comment to your answer, but a contribition to your list of errata in "Topology and Groupoids": https://math.stackexchange.com/q/4559806 – Paul Frost Oct 24 '22 at 22:52