Intereting Posts

Polycyclic groups and Group extension
If $f$ and $g$ are integrable, then $h(x,y)=f(x)g(y)$ is integrable with respect to product measure.
Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$.
A non-abelian group such that $G/z(G)$ is abelian.
Algebraic vs. Integral Closure of a Ring
How to calculate the integral of $x^x$ between $0$ and $1$ using series?
Minimum size of the generating set of a direct product of symmetric groups
Prove that $\frac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdots\sqrt{2}$
Please integrate $\int \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx$
0.246810121416…: Is it a algebraic number?
Does a countable set generate a countable group?
Riemann's thinking on symmetrizing the zeta functional equation
Original Papers on Singular Homology/Cohomology.
Fourier transform in $L^p$
Generalized Second Borel-Cantelli lemma

The following question is from my algebraic topology exam which I was unable to solve.

Let $X$ be a orientable connected closed $n$-manifold. Let $ p \in X$. Show that $H^n (X-{p},R)=0$, where $R$ is some ring.

I observed that since $X$ is orientable so $X-p$ is also orientable hence $H_n(X-p,R) \cong R$. I think that Poincare duality may help us to prove this result but I am unable to prove it. Any ideas?

- Homotopy equivalence from torus minus a point to a figure-eight.
- Finitely generated singular homology
- How to understand the Todd class?
- CW construction of Lens spaces Hatcher
- Is the center of the fundamental group of the double torus trivial?
- Properties of Surgery on Manifolds?

- Is totally disconnected space, Hausdorff?
- Passage to fixed point spaces is object function of a contravariant functor?
- Quotient space and Retractions
- Local coefficients involved in the obstruction class for a lift of a map
- Second Stiefel-Whitney Class of a 3 Manifold
- Is there a “geometric” interpretation of inert primes?
- What is this space homeomorphic/homotopy equivalent to?
- Why does «Massey cube» of an odd element lie in 3-torsion?
- Understanding construction of open nbds in CW complexes
- Can we think of a chain homotopy as a homotopy?

There is the long exact sequence in homology of the pair $(X, X \setminus p)$ (everything is with $\mathbb{Z}$-coefficients unless indicated otherwise):

$$0 \to H_n(X \setminus p) \to H_n(X) \to H_n(X, X \setminus p) \to H_{n-1}(X \setminus p) \to H_{n-1}(X) \to 0$$

(recall that the local homology groups satisfy, by excision, $H_n(X, X \setminus p) = \mathbb{Z}$ and $H_k(X, X \setminus p) = 0$ for $k \neq n$).

The fundamental class $[X] \in H_n(X)$ is sent to a generator of the local homology group $H_n(X, X \setminus p) = \mathbb{Z}$ (either by definition, or cf. Hatcher, Theorem 3.26 for example). Hence $H_n(X) \to H_n(X, X \setminus p)$ is an isomorphism and thus $H_n(X \setminus p) = 0$ and $H_{n-1}(X \setminus p) \cong H_n(X)$.

The universal coefficient theorem reads

$$H^n(X \setminus p; R) \cong \operatorname{Hom}_{\mathbb{Z}}(H_n(X \setminus p), R) \oplus \operatorname{Ext}^1_{\mathbb{Z}}(H_{n-1}(X \setminus p), R).$$

We already know the first summand vanishes. By Poincaré duality and universal coefficients, $H_{n-1}(X) \cong H^1(X) \cong \hom(H_1(X), \mathbb{Z})$. Thus $H_{n-1}(X)$ is torsion-free, and it is also finitely generated because $X$ is compact, hence it is free as an abelian group (as $\mathbb{Z}$ is a PID). So the second summand vanishes too ($H_{n-1}(X \setminus p) = H_{n-1}(X)$ is free) and finally $H^n(X \setminus p) = 0$.

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