# Problem of the Week #284 - Jan 07, 2019

Status
Not open for further replies.

#### Euge

##### MHB Global Moderator
Staff member
Happy New Year, everyone! Here's this week's problem!

-----
If $M$ and $N$ are topological manifolds with boundary of dimensions $m$ and $n$, respectively, show that $M \times N$ is an $(m+n)$-manifold with boundary $\partial M \times N \cup M \times \partial N$.

-----

Note $(x,y) \in \partial(M \times N)$ if and only if the relative homology $H_{m+n}(M\times N, M\times N \setminus \{(x,y)\}) = 0$. Since $M\times N \setminus \{(x,y)\} = (M\setminus \{x\}) \times N \cup M \times (N \setminus \{y\})$, the relative Künneth formula gives $$H_{m+n}(M\times N, M\times N \setminus \{(x,y)\}) \approx \bigoplus_{i + j = m + n} H_i(M,M\setminus \{x\}) \otimes H_j(N,N\setminus \{y\}) \approx H_m(M,M\setminus\{x\}) \otimes H_n(N,N\setminus\{y\})$$ Both $H_m(M,M\setminus\{x\})$ and $H_n(N,N\setminus\{y\})$ are zero or infinite cyclic, so the left-hand side is trivial if and only if $H_m(M,M\setminus\{x\}) = 0$ or $H_n(N,N\setminus \{y\}) = 0$, i.e., $(x,y)\in \partial M \times N \cup M\times \partial N$. Hence $\partial(M\times N) = \partial M \times N \cup M \times \partial N$.