TITLE: Existence of a path between two sides of a square, which is contained in the intersection of two open subsets of it
QUESTION [19 upvotes]: Denote by $I$ the interval $[0,1]$. Let $A,B$ be two open subsets of the square $I\times I=I^2$ such that $A\cup B=I^2$, $\{0\}\times I\subseteq A$ and $\{1\}\times I\subseteq B$. In more geometric terms, the left edge of the square is contained in $A$ and the right edge is contained in $B$.
My question is: does there exist a continuous path $\gamma:I\to A\cap B$ such that $\gamma(0)\in I\times\{0\}$ and $\gamma(1)\in I\times\{1\}$? In other words, is there a curve connecting the two horizontal sides of the square, whose trace is contained in $A\cap B$ ?

REPLY [2 votes]: The cleanest proof of this I know is to use a Mayer-Vietoris sequence. You can do directly on a square using relative homology, but it's a bit cleaner to note that the original statement is equivalent to asking that for any pair of open subsets $A$ and $B$ of the cylinder $S^1\times I$ such that $A$ contains $S^1\times \{0\}$ and $B$ contains $S^1\times \{1\}$, there exists some loop $A\cap B$ which is not contractible in $S^1\times I$. You can get this equivalence by gluing the top and the bottom of the square together and adding in a strip along this glue line which is in both $A$ and $B$. There's some little details to take care of, but it's possible to ensure that the subspace of the cylinder outside of the extra strip in both $A$ and $B$ is just your original square with its original subsets. Once you have your non-trivial loop, you can find that it must, at some point, traverse over the part of the cylinder that is your original square - which gives you your answer.
That taken care of, we can just write out one part of the Mayer-Vietoris sequence for $A$ and $B$:
$$H_1(A\cap B)\overset{f}{\rightarrow} H_1(A)\oplus H_1(B) \overset{g}{\rightarrow} H_1(S^1\times I)$$
where the first map $f$ is the sum of the two embeddings $A\cap B\rightarrow A$ and $A\cap B\rightarrow B$ and where the second map $g$ takes $h_1\oplus h_2$ to the difference of the embeddings of $h_1$ and $h_2$ into $H_1(S^1\times I)$.
The image of this first map must equal the kernel of the second. However, if we let $\alpha\in H_1(A)$ be the loop around $S^1\times \{0\}$ and $\beta\in H_1(B)$  be the similarly oriented loop around $S^1\times \{1\}$, it's clear that these two loops embed to the same thing in $H_1(S^1\times I)$, hence $\alpha\oplus \beta \in \ker g$. As a result, there is some class $h\in H_1(A\cap B)$ such that the embedding of $h$ into $H_1(A)$ gives $\alpha$. Clearly, any chain in the homology class of $h$ contains some loop which is non-trivial in $S^1\times I$, so we are done.