(Hatcher 3.B.4) Show that the cross product of fundamental classes for closed $R$-orientable manifolds $M$ and $N$ is a fundamental class for $M\times N$. Assume $\dim M = m,\dim N = n$.
Let $[M]$ and $[N]$ be fundamental classes of $M$ and $N$ respectively. Since $M,N$ are closed orientable manifolds, by Kunneth formula,
$$H_{m+n}(M\times N;\Bbb Z)\simeq H_{m}(M;\Bbb Z)\otimes H_n(N;\Bbb Z)$$
and this isomorphism is given by (homology) cross product. Hence, $[M]\times [N]$ is a generator of $H_{m+n}(M\times N;\Bbb Z)$. Now I need to check if $H_{m+n}(M\times N;\Bbb Z)\to H_{m+n}(M\times N-\{(x,y)\};\Bbb Z)$ maps $[M]\times[N]\to \mu_{(x,y)}$ where $\mu_{(x,y)}$ is a local orientation of $M\times N$.
But btw, what is a local orientation of $M\times N$? Can I express it as local orientations of $M$ and $N$? I have no idea how to proceed further. Could you help?
Note. One might remind Hatcher exercise 3.3.5 ($M\times N$ is orientable if and only if $M$ and $N$ are both orientable). I don't know the proof using direct argument using definition (I think it's quite complicated) but I know the proof using orientable cover which does not give local orientation of $M\times N$ (I remember there is a post asking this question).
Best Answer
Theorem 1: Let $A\subseteq_\text{open} X,\ B\subseteq_\text{open} Y$. Consider the realtive chain complexes $\mathcal C=C_*(X,A)$ and $\mathcal C'=C_*(Y,B)$. Then there is a natural split short exact sequence $$0\to \bigoplus_{p+q=n}H_p(X,A)\otimes H_q(Y,B)\xrightarrow{\text{cross product}} H_n(X\times Y,A\times Y\cup X\times B)\to \bigoplus_{p+q=n-1}\text{Tor}\big(H_p(X,A),H_q(Y,B)\big)\to 0$$
Theorem 2: For a manifold $X$ of dimension $d$ and $x\in X\backslash \partial X$ we have $H_n(X,X\backslash x)=\begin{cases} \Bbb Z &\text{ if }n=d,\\ 0 &\text{ otherwise.}\end{cases}$
Proof: Use excision theorem considering a closed coordinate ball $\overline B\cong \{z\in \Bbb R^d:|z|\leq 1\}$ with $x\in \text{int}(\overline B)$.
Corollary 1: For two manifolds $M,\ \dim M=m$ and $N,\ \dim N=n$ with $x\in M\backslash \partial M,\ y\in N\backslash \partial N$ we have the isomorphism $$H_m(M,M\backslash x)\otimes H_n(N, N\backslash y)\xrightarrow[\cong]{\text{cross product}} H_{m+n}\big(M\times N,M\times N\backslash (x,y)\big)$$
Proof: Combine Theorem 1 and Theorem 2.
Observation 1: For two manifolds $M,\ \dim M=m$ and $N,\ \dim N=n$ with $x\in M\backslash \partial M,\ y\in N\backslash \partial N$ and orientations $\{\mu_p:p\in M\backslash \partial M\},\ \{\nu_q:q\in N\backslash \partial N\}$ we have the following commutative diagram
where as usual $x\in U\subseteq_\text{open}M$ and $\mu_U\in H_m(M, M\backslash U)\cong \Bbb Z$ is a generator such that $\mu_U\longmapsto \mu_p$ for each $p\in U$ under the inclusion induced map $H_m(M,M\backslash U)\to H_m(M,M\backslash p)$. Similarly, define $\nu_V$.
The bottom arrow is an isomorphism due to Corollary 1. A similar reasoning(cross-product argument) says the top arrow is also an isomorphism. The left arrow is an isomorphism due to the choice of $U$ and $V$ while defining orientations. Now, the commutativity of the square implies, the right arrow is an isomorphism.
Theorem 3: Let $M,N$ be two compact oriented manifolds with the fundamental classes $[M]\in H_m(M,\partial M),\ [N]\in H_n(N,\partial N)$.
Then, the fundamental class $[M\times N]\in H_{m+n}\big(M\times N, \partial(M\times N)\big)$ is same as the $[M]\times [N]$.
Proof: Consider $x\in M\backslash \partial M, y\in N\backslash \partial N$, then $(x,y)\in M\times N\backslash \partial(M\times N)$. So, the inclusion induce map $H_{m+n}\big(M\times N,\partial(M\times N)\big)\longrightarrow H_{m+n} \big(M \times N, M \times N \backslash (x, y)\big)$ is an isomorphism. Now, consider the commutative diagram below.
The bottom arrow is an isomorphism due to Corollary 1. The left arrow is isomorphism as $x\in \text{int}(M),\ y\in \text{int}(N)$. So, the top arrow is an isomorphism, i.e., the cross product map $H_m (M, \partial M ) \otimes H_n (N, \partial N )\xrightarrow{\times}H_{m+n} \big(M \times N, \partial(M \times N )\big)$ sends $[M]\otimes [N]$ to $[M\times N]$.