4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ As long as I remember, the Alexander duality is for spheres while Pontryagin is for manifolds is more general. $\endgroup$ – Wlod AA Sep 4 '18 at 6:09
  • $\begingroup$ @WlodAA Concerning Alexander-Pontryagin duality I used the info fromhttps://www.encyclopediaofmath.org/index.php/Alexander_duality $\endgroup$ – Taras Banakh Sep 4 '18 at 7:27
  • $\begingroup$ @WlodAA Pontrjagin duality is about locally compact abelian groups. Are you thinking of Poincaré duality? $\endgroup$ – Arun Debray Sep 4 '18 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 Sep 4 '18 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 Sep 4 '18 at 17:41
3
$\begingroup$

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 $H^{n-2}(B)=0$).

$\endgroup$
  • $\begingroup$ 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 Sep 4 '18 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 Sep 4 '18 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.