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This is an exercise 3.3.5 in Hatcher's Algebraic Topology book.

3.3.5: Show that $M \times N$ is orientable if and only if $M$ and $N$ are both orientable.

Notation: For $x \in M$, $H_n(M|x):=H_n(M,M-x)$.

Definition: An orientation of an $m$-dimensional manifold $M$ is a function $x \to \mu_x$ assigning to each $x \in M$ a local orientation of $\mu_x \in H_n(M|x)$ satisfying the local consistency condition that each $x \in M$ has a neighbourhood $U \approx \mathbb R^m \subset M$ containing an open ball $B$ of finite radius about $x$ such that all the local orientations $\mu_y$ at points $y \in B$ are the images of one generator $\mu_B$ of $H_n(M|B) \approx H_n(\mathbb R^m |B)$ under the natural maps $H_n(M|B) \to H_n(B|y)$ induced by inclusion. If an orientation exists for $M$, then $M$ is called orientable.

Attempt: Let $x \to \mu_x, \ y \to \mu_y$ be respective functions assigning to $x\in M$ and $y\in N$ local orientations as described above.

By Section 3B of Hatcher, there is a cross product homomorphism $$H_i(M,M-x) \otimes H_j(N, N-y) \to H_{i+j}(M\times N, (M-x)\times (N-y)).$$

Since $M-x\subset M$ and $N-y \subset N$, we have $(M-x)\times (N-y) \subset M \times N$. At this point, can we say that $$(M-x)\times (N-y) \subset M \times N-(x,y)?$$

If this holds, then the function $(x,y) \to \mu_x \otimes \mu_y$ assigning to $(x,y) \in M\times N$ local orientation in the above manner gives the global orientation for $M\times N$.

Any suggestion will be appreciated.

1 Answers1

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Checking whether $(M-x) \times (N-y) \subset (M \times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) \times (N-y)$ also excludes $\{x\} \times N$ and $M \times \{y\}$.

If you look at page $276$ of Hatcher, you see that $$C_\ast(M)/C_\ast(M-x) \otimes C_\ast(N)/C_\ast(N-y) \cong C_\ast(M \times N)/C_\ast([(M-x) \times N ]\cup [M \times (N-y)])$$ The latter term is the same as $C_\ast(M\times N)/C_\ast(M\times N - (x,y))$ and should help give the isomorphism in homology you're looking for.

Osama Ghani
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  • The latter term is the same as $C_\ast(M\times N)/C_\ast(M\times N - (x,y))$ due to a set theoretical identity? –  Dec 22 '18 at 17:52
  • Yes! Hopefully it's clear that $(M \times N) - (x,y) \subset [(M-x) \times N] \cup [M \times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x \times N) \cap (M \times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan). – Osama Ghani Dec 22 '18 at 17:57