I'm trying to prove (without results) that if $M$ is an orientable topological manifold then every open set of $M$ is orientable. I'm assuming as definition of orientation the reformulation 3 of orientation.
Homological orientation of a manifold
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algebraic-topology
manifolds
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3Hint: Local homology has its name because $H_n(M,M \setminus x) = H_n(U, U \setminus x)$ for any neighborhood of $U$ of $x$. Here I write equality for the natural isomorphism induced by excising $U^c$. – 2017-02-19
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0Yes, in this way I have a generator $M_x\in H_n(U,U\setminus x)$, but how I can check that the choice is continuos? – 2017-02-20
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0The *collection* of generators was continuous to begin with. First try showing that you can shrink the neighborhood in the definition of continuity, since you will need a neighborhood in $U$ ($U$ in earlier comment, not the $U$ on the linked page.) – 2017-02-20