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When working with smooth manifolds, $M^m$ and $N^n$, it is straightforward to see how orientations at points $p\in M$ and $q\in N$ (i.e. ordered bases for the tangent spaces) give rise to an orientation at the point $(p,q) \in M \times N$, given a convention about which basis should come first. However in the continuous setting, it isn't so clear to me how a choice of generators for $H_m(M, M-p)$ and $H_n (N, N-q)$ should naturally give rise to a generator for $H_{n+m} (M \times N, M\times N -(p,q))$. I think you can probably do it by the same method as in the smooth case once you choose local coordinates and a convention about how ordered bases for $\mathbb{R}^n$ correspond to generators of $H_n(\mathbb{R}^n, \mathbb{R}^n-0)$, but I was hoping for some purely algebraic-topology construction that doesn't use tangent vectors. Thank you for your time.

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The answer comes from the Künneth formula. Of course, by excision, you are localizing and looking at $H_m(D^m,\partial D^m)$, $H_n(D^n,\partial D^n)$, and $H_{m+n}(D^{m+n},\partial D^{m+n})$. An orientation, as you pointed out, is a choice of generator for $H_m(D^m,\partial D^m) \cong \mathbb Z$, etc. Now, the Künneth isomorphism [see, e.g., Hatcher p. 276 for the relative statement] \begin{align*} H_m(D^m,\partial D^m)\otimes H_n(D^n,\partial D^n) &\overset{\cong}{\to} H_{m+n}(D^m\times D^n,\partial D^m\times D^n \cup D^m\times\partial D^n)\\ &\cong H_{m+n}(D^{m+n},\partial D^{m+n}) \end{align*} gives you what you want. (It's probably easiest to visualize the last step thinking of cubes, rather than disks.)

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ah, of course. Thank you, Ted –  Pliny the ill Jul 11 '13 at 21:33
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