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I am struggling to understand the proof of the Van Kampen theorem in Hatcher's book, "Algebraic Topology", that one can find at page 44 of https://pi.math.cornell.edu/~hatcher/AT/AT.pdf.

In particular, I am not able to answer the following two questions:

1) In order to prove the injectivity of the map $\Phi: *_\alpha\ \pi(A_\alpha)\to\pi(X)$, the author requires the triple insersection $A_\alpha\cap A_\beta\cap A_\gamma$ to be path connected, while other proofs of this theorem do not require it. Is this really needed?

2) Is there a way to visualize what is happening in the figure at page 45? For example he says that one has to shift the squares in the middle row in order to avoid intersection with four open sets. Why is this achieving what he wants?

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2 Answers 2

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1) "...while other proofs of this theorem do not require it."

This seems unlikely because the theorem is false without the triple-intersection assumption. Perhaps you are thinking of the statement where $X$ is covered by two open sets, $U$ and $V$. In that case the triple intersections are only

$U \cap U \cap U = U,$

$U \cap U \cap V = U \cap V,$

$U \cap V \cap V = U \cap V,$

$V \cap V \cap V = V,$

which you assumed were already connected. So the triple-intersection condition was tautologically satisfied when the open cover is only two open sets.

2) You imagine the bottom edge and the top edge as being two different factorizations of the map $f$, and the entire square is a homotopy between them (constant on the right and left sides), where each rectangle lies in some $U_\alpha$. We fiddle with this diagram a bit so that each point lies in exactly three rectangles. Because each of these rectangles are chosen to lie in some specified $U_\alpha$, that means that every point lies in some specified triple-intersection. If the vertical lines didn't have 'breaks', then you would instead have points that live in 4 different rectangles, so most naturally live in some quadruple-intersection.

Now one shows inductively that you can take the bottom edge and move it to the top edge by changing one small part of the curve at a time (eg, changing from the bottom-right edges of a rectangle to the top-left edges), and that in doing so, you change $f$ by an element of the group we think is the kernel of the map $\Phi$.

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  • $\begingroup$ 1) I will surely check the precise assumptions of the other proofs. 2) "We fiddle with this diagram a bit so that each point lies in exactly three rectangles. Because each of these rectangles are chosen to lie in some specified Uα, that means that every point lies in some specified triple-intersection." I am confused about this last statement: from what I understand, the rectangles are not necessarily mapped in different Uα, so wouldn't it be more correct to say that every point lies AT MOST in some triple intersection? Sorry if it is a silly question, I am just a Physicist ;) $\endgroup$
    – samario28
    Commented Jan 26, 2020 at 15:08
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  1. In page 44, above the proof of the theorem, there is an explanation about the triple-intersection assumption. The theorem fails to hold without this assumption. Hatcher's van Kampen theorem is more general than other books, because other books usually state the van Kampen theorem using only two open sets.

  2. The perturbation trick can be achieved via the tube lemma. The trick is needed because the intersection of four $A_\alpha$'s need not be path-connected, so we are using the trick to make each vertex lie in at most three rectangles. This is explained in the proof, in the last paragraph of page 45.

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  • $\begingroup$ 1. Thank you for the suggestion. 2. I will have to look for this tube lemma that you cite. $\endgroup$
    – samario28
    Commented Jan 26, 2020 at 15:09

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