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The only proof I know involves the Mayer - Vietoris sequence. Is there an elementary proof?

Asvin
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  • I am very rusty, but isn't the intersection of 2 path-connected sets path-connected almost by definition? – gnometorule Mar 25 '15 at 04:08
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    @gnometorule: Hint: consider two circular arcs – shalin Mar 25 '15 at 04:11
  • @Shalop: Thanks. – gnometorule Mar 25 '15 at 04:12
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    I think you can also prove this using a groupoid version of Seifert-van Kampen. – Qiaochu Yuan Mar 25 '15 at 04:18
  • I do not believe this statement in absence of extra assumption ($A,B$ open or whatever). Take $A=D^+\cup ]0,1/3]\cup [2/3,1[$ and $B=D^-\cup ]0,1[$ where $D^+$ is the open upper half-disk ($|z|<1,\ \Im(z)>0$ and $D^-$ is the open lower half-disk ($|z|<1,\ \Im(z)<0$. Then $A,B$ are path connected, $A\cup B=D$ (the full disk) is simply connected but $A\cap B=]0,1/3]\cup [2/3,1[$ is not connected. Would you mind elaborating ? – Duchamp Gérard H. E. Mar 25 '15 at 10:15
  • You are of course correct, I was operating under the assumption that A and B are open @DuchampGérardH.E. I am pretty sure the proof given as an answer goes through in that case, I am not sure how much more general A and B could be. – Asvin Mar 25 '15 at 10:37
  • @Asvin OK point taken ! – Duchamp Gérard H. E. Mar 25 '15 at 10:46
  • @Asvin Suggestion of improvement : I cannot prevent thinking it would be wise to tag your question officially (and not only between brackets) "algebraic topology" in order to avoid the curious but external reader the "club effect" of hypotheses contextual to a domain. Or, if you want to reach a wider community, make additions. – Duchamp Gérard H. E. Mar 25 '15 at 11:49
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    Also see http://math.stackexchange.com/questions/1163650/if-mathbbrn-u-cup-v-for-path-connected-u-v-then-u-cap-v-is-path-connec – Stefan Hamcke Mar 25 '15 at 14:28

2 Answers2

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Let $x, y \in A \cap B$. Let $\alpha : [0, 1] \rightarrow A$ be a path from $x$ to $y$, and let $\beta : [0, 1] \rightarrow B$ be a path from $x$ to $y$. Let $H : [0, 1] \times [0, 1] \rightarrow A \cup B$ be a homotopy of $\alpha$ to $\beta$ with fixed endpoints. Divide $[0, 1] \times [0, 1]$ into subsquares so tiny that each subsquare gets mapped by $H$ entirely to $A$ or to $B$ (i.e. consider the covering of the square given by $\{H^{-1}(A), H^{-1}(B)\}$ and take its Lebesgue number $\epsilon > 0$, then divide the square $[0, 1] \times [0, 1]$ into subsquares of diagonal $< \epsilon$). Now color each of the subsquares blue if gets mapped entirely to $A$, red if it gets mapped entirely to $B$, and black if it gets mapped entirely to $A \cap B$. It suffices to convince yourself that, starting at the left side of the square, you can travel to the right side of the square, passing only through edges that bound either a black square or two squares, one being red and the other being blue. Since endpoints are fixed in the homotopy, this will give you a path from $x$ to $y$ that lies in $A \cap B$.

Pedro
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As Qiacho comments, the result does follow from the groupoid version of the Seifert-van Kampen Theorem, which dates from 1967, is exploited in the book Topology and Groupoids, and discussed on this mathoverview question and answer. This paper gives a relevant small correction to one part of the book.

In effect, Pedro is running through part of the proof of that theorem.

Community
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Ronnie Brown
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