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Let $n\ge 3$ and $X$ be a compact connected $n$-manifold (without boundary).

I need a reference to the following facts (which I believe are true at least in dimension $n=3$):

Fact 1. For every closed connected subset $A\subset X$ that can be embedded to $\mathbb R^{n-1}$ the complement $X\setminus A$ is connected.

Fact 2. For any closed subset $B\subset X$ whose connected components can be embedded to $\mathbb R$, the identity embedding $X\setminus B\to X$ induces an injective homomorphism $H_1(X\setminus B;G)\to H_1(X;G)$ in singular homologies with coefficients in some group $G$ (for example $\mathbb Z$ or $\mathbb Z/2\mathbb Z$).

Remark. The Alexander-Pontryagin Duality Theorem implies that Facts 1 and 2 are true if $X$ is the $n$-sphere. So, I need these facts for an arbitrary compact connnected $n$-manifold without boundary.

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    $\begingroup$ As long as I remember, the Alexander duality is for spheres while Pontryagin is for manifolds is more general. $\endgroup$
    – Wlod AA
    Commented Sep 4, 2018 at 6:09
  • $\begingroup$ @WlodAA Concerning Alexander-Pontryagin duality I used the info fromhttps://www.encyclopediaofmath.org/index.php/Alexander_duality $\endgroup$
    – Taras Banakh
    Commented Sep 4, 2018 at 7:27
  • $\begingroup$ @WlodAA Pontrjagin duality is about locally compact abelian groups. Are you thinking of Poincaré duality? $\endgroup$
    – Arun Debray
    Commented Sep 4, 2018 at 13:15
  • $\begingroup$ Use a long exact sequence (for example, mathoverflow.net/questions/124816/…) and Poincaré duality with compact supports (for example, Theorem 3.35 of Hatcher). $\endgroup$
    – Chris Gerig
    Commented Sep 4, 2018 at 17:35
  • $\begingroup$ @ArunDebray Encyclopedia of Math. (encyclopediaofmath.org/index.php/Alexander_duality) writes Alexander-Pontryagin duality. The Poincare duality is something different. $\endgroup$
    – Taras Banakh
    Commented Sep 4, 2018 at 17:41

1 Answer 1

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Restating my comment above (which linked to another MO post):

For an open subspace $U\subset X$ there is a long exact sequence (via the normal LES for the pair $(X,X-U)$ and excision) $$ \cdots\to H^\ast_c(U) \to H^\ast_c(X) \to H^\ast_c(X-U) \to H^{\ast+1}_c(U)\to\cdots$$ and Poincaré duality with compact supports $H^\ast_c(M)\cong H_{\dim M-\ast}(M)$, see for example Theorem 3.35 of Hatcher's bible.

Apply $\ast=n$ for Fact 1 (using $\dim A\le n-1$) and $\ast=n-1$ for Fact 2 (using $H^{n-2}(B)=0$).

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  • $\begingroup$ Sorry, but I cannot understand how do you use the excision for identifying $H_c^*(U)$ with $H^*_c(X,X-U)$ in LES. The latter cohomology group should rather be identified with $H^*(X/(X-U))$ (at least for nice $X-U$)? $\endgroup$
    – Taras Banakh
    Commented Sep 4, 2018 at 21:06
  • $\begingroup$ By definition, $H^\ast_c(U):=\lim H^\ast(U,U-K)$ exhausting $U$ by compact subsets $K$. Then excision identifies this with $\lim H^\ast(X,X-K)$, still over $K\subseteq U$, which should be the desired relative cohomology. $\endgroup$
    – Chris Gerig
    Commented Sep 4, 2018 at 22:32

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