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Sorry I have to quote a major part of content from Bredon’s /Topology and Geometry/ as a screenshot. In the following definition of the basis $U_\alpha$, why using the closure of $U$ but not $U$ directly? It seems the arguments for $U_\alpha$ being a basis still hold with $j_{x, U}$.

Thanks!

enter image description here

edit: I added another screenshot on the proof of $U_\alpha$ being a basis. It may require taking the closure of $U$ somewhere. But I cannot find it.

enter image description here

onRiv
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    I don’t know the specific reason, but I recall that 1) $M-A$ is again a manifold, if $A$ is closed. I don’t know whether this is used, but it is certainly nice to have. 2) depending on the version of excission Bredon proves, there might be some closure constraints involved. – Jonas Linssen Mar 01 '22 at 10:43
  • @JonasLinssen Thanks. I am checking Bredon’s version of the excision theorem. I haven’t seen it before. I am coming across the topology from Hatcher’s book (sec3.3 about the Poincare Duality), where the topology on the orientation bundle defined a little unclear and there is a quite technical lemma to follow. Any other materials about the orientation of topological manifolds can you suggest please? – onRiv Mar 01 '22 at 11:12
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    So your question is why Bredon does not use $\alpha \in H_n(M,M-U;G)$? – Paul Frost Mar 01 '22 at 12:10
  • @PaulFrost Yeah He uses the map: $j_{x, \bar{U}}: H_n(M, M- \bar{U}; G) \rightarrow H_n(M, M-x; G)$, and take the image of each map on $\alpha$ as a basis. I am wondering which part requires considering the $j_{x, \bar{U}}$ but not $j_{x,U}$ – onRiv Mar 01 '22 at 12:14

1 Answers1

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Your are right.

For all $R \subset S \subset M$ let us denote by $j_{R,S} : H_n(M,M-S;G) \to H_n(M,M-R;G)$ the map induced from inclusion. Then $j_{x,A} = j_{\{x\},A}$ (this notation can be used also for non-closed $A$). Let $j_U = j_{U, \overline U} : H_n(M,M-\overline U;G) \to H_n(M,M- U;G)$.

Let us define $U'_\beta = \{j_{x,U}(\beta) \mid x \in U \}$ for all $\beta \in H_n(M,M- U;G)$. For $\alpha \in H_n(M,M-\overline U;G)$ we then have $$U_\alpha = U'_{j_U(\alpha)} .$$ Thus the set $\mathcal U = \{U_\alpha \mid U \subset M \text{ open, } \alpha \in H_n(M,M-\overline U;G)\}$ is a subset of $\mathcal U' = \{U'_\beta \mid U \subset M \text{ open, } \beta \in H_n(M,M- U;G)\}$.

Bredon has proved in Proposition 7.3 that $\mathcal U$ is the basis for a topology on $\Theta \otimes G$. We shall prove (by modifying Bredon's arguments) that also $\mathcal U'$ is the basis for a topology on $\Theta \otimes G$ and the topologies generated by $\mathcal U, \mathcal U'$ agree.

To do so, we shall show that for any two $U'_\alpha, V'_\beta \in \mathcal U'$ and each $\tilde x \in U'_\alpha \cap V'_\beta$ we can find an open neigborhood $W$ of $x$ and $\gamma \in H_n(M,M-\overline W;G)$ such that $\tilde x \in W_\gamma \subset U'_\alpha \cap V'_\beta$.

Since $W_\gamma \in \mathcal U \subset \mathcal U'$, we conclude that $\mathcal U'$ forms a basis. Moreover, taking $U = V$ and $\alpha = \beta$, we see that each $U'_\alpha$ is the union of suitable $W_\gamma$. This shows the topologies generated by $\mathcal U, \mathcal U'$ agree.

So let us consider $\tilde x \in U'_\alpha \cap V'_\beta$. The element $\tilde x$ lies in a unique $\Theta_x \otimes G$. We thus have $\tilde x = j_{x,U}(\alpha) = j_{x,V}(\beta)$. We can find a convex open set $W$ in a euclidean neigborhood of $x$ in $M$ such that $\overline W \subset U \cap V$. Then $j_{x,\overline W}$ is an isomorphism. Let $\gamma \in H_n(M,M - \overline W;G)$ be the unique element such that $j_{x,\overline W}(\gamma) = \tilde x$. In particular $\tilde x \in W_\gamma$.

The homomorphism $j_{\overline W,U} : H_n(M,M-U) \to H_n(M,M-\overline W)$ has the property $j_{\overline W,U}(\alpha) = \gamma$ because $j_{x,\overline W}(\gamma) = \tilde x = j_{x,U} (\alpha) = j_{x,\overline W}(j_{\overline W,U}(\alpha))$. Now let $\tilde y \in W_\gamma$. We have $\tilde y = j_{y,\overline W}(\gamma)$ for a unique $y \in W$. Thus $\tilde y = j_{y,\overline W}(j_{\overline W,U}(\alpha)) = j_{y,U}(\alpha) \in U'_\alpha$. Similarly $\tilde y \in V'_\beta$.

Paul Frost
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  • Thanks! This is a very clear explanation. – onRiv Mar 01 '22 at 13:44
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    @onriv I found that a slight modification of Bredon's proof shows that you can indeed take $U$ instead of $\overline U$. I made an update of my answer. But I do not think that the basis $\mathcal U'$ has any advantage over $\mathcal U$. – Paul Frost Mar 01 '22 at 23:32
  • originally I was vaguely taking only $U_\alpha$ as a basis for U that is homeomorphic to some small open ball. That seems a lot smaller than Bredon’s definition. Reading your answer I realized that I in fact doesn’t gain a solid understanding in Proposition 7.1. The proof of it seems not so oblivious and involving a version of homotopy invariant for homotopic equivalent pairs (X, A) and (Y, B). Using this version of homotopic invariant it seems possible to prove $j_{U, \bar{U}}$ is isomorphic for small open ball U and then make the original question I asked more easier to see? – onRiv Mar 02 '22 at 11:11
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    @onriv It sufffices indeed to to take any basis $\mathcal B$ for $M$ and then take the $U_\alpha$ (or $U'\beta$) with $U \in \mathcal B$ as a basis for $\Theta \otimes G$. Now you can take for $\mathcal B$ the set of all convex open sets with compact closure contained in a euclidean neigborhood. This gives a smaller basis than Bredon's. But I am not sure that $j{U, \overline U}$ is an isomorphism for "small" $U$: – Paul Frost Mar 02 '22 at 11:45