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Suppose that we can write

$$ \mathbb{R}^n=U\cup V, $$ where $U,V$ are open and path-connected. It is easy to show using Mayer-Vietoris sequence that $U\cap V$ is path-connected as well.

Is there a proof which does not use homology, ideally something just using point-set topology?

Jimmy R
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    Nice question, +1. ${}$ Since the proof via the Mayer-Vietoris sequence uses the fact that $\tilde H_1(R^n)=0$, an elementary proof would use the simply-connectedness of $R^n$, so it should involve a map $D^2\to R^n$ extending the union of two paths from $x$ to $y$, one of which lies in $U$ and the other one in $V$. – Stefan Hamcke Feb 24 '15 at 19:09

1 Answers1

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This fact is a Corollary of the many pointed version of the Seifert-van Kampen theorem, published in 1967.

If $X= U \cup V$ is path connected, where $U,V$ are open and path connected, and $U \cap V$ has $n$ path components, then the fundamental group of $X$ has the free group $F$ on $n-1$ generators as a retract. See Topology and Groupoids Section 8.4; see also this paper on the Phragmen-Brouwer Property, which gives a small correction to that section, and further reference to the literature. This paper proves by groupoid methods that if instead $U,V,W=U \cap V$ have $n_U,n_v,n_W$ path components, then the rank of the free group $F$ is $n_W-n_U-n_V+1$.

Ronnie Brown
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