Why is the one-point compactification of $S^5\times \mathbb{R}$ not a manifold?

Question

Why is the one-point compactification of $S^5\times\mathbb{R}$ not a manifold? This problem was on qualifying exam, but I could not prove it.

By simple drawings, one can see that the one-point compactification of $S^0\times \mathbb{R}$ and $S^1\times \mathbb{R}$ are homotopic to $S^1\vee S^1$ and $S^2\vee S^1$ respectively. This leads me to guess that the one-point compactification of $S^5\times \mathbb{R}$ might be homotopic to $S^6\vee S^1$, which is clearly not a manifold because of its wedge point.

However, how do I actually prove it? Moreover, there was a hint that if $X$ is the one-point compactification of $S^5\times\mathbb{R}$ and if it is an $n$-manifold, then $H_k(X,S^5\times\mathbb{R})\cong H_k(D^n,S^{n-1})$ by the excision, and this leads to a contradiction. Why?

Thank you in advance!

Answer 1

First, note that it's not sufficient to prove that $X$ is homotopy equivalent to $S^6 \vee S^1$ in order to prove that it's not a manifold. Indeed you could "thicken" $S^6 \vee S^1$ and get an 8-manifold (if you want, imagine figure eight $S^1 \vee S^1$ embedded in the plane $R^2$, and consider a small neighborhood of that: that's a 2-manifold; the idea is the same). "Being a manifold" is not preserved by homotopy equivalence.

Now suppose that $X = S^5 \times \mathbb{R} \cup \{ \infty \}$ were an $n$-manifold. Then by definition, $\infty \in X$ must have a neighborhood $U \subset X$ homeomorphic to $D^n$, where $\infty$ corresponds to $0 \in D^n$. Using excision, you get $$H_k(X, S^5 \times \mathbb{R}) \cong H_k(U, U \setminus \{ \infty \})$$ and since homology is invariant under homeomorphisms, you get: $$H_k(X, S^5 \times \mathbb{R} \cup \{ \infty \}) \cong H_k(D^n, D^n \setminus \{0\}) \cong \begin{cases} \mathbb{Z} & \text{if } k = n; \\ 0 & \text{if } k \neq n. \end{cases}$$

Now, clearly any point in $X \setminus \{\infty\}$ has a neighborhood homeomorphic to $D^6$ (because $S^5 \times \mathbb{R}$ is a $6$-manifold), thus $n = 6$. If you write down the long exact sequence in homology of the pair $(X, S^5 \times \mathbb{R})$, you get: $$H_{k-1}(X, S^5 \times \mathbb{R}) \to H_k(S^5) \to H_k(X) \to H_k(X, S^5 \times \mathbb{R}) \to H_{k+1}(S^5 \times \mathbb{R})$$ The homology of $S^5 \times \mathbb{R} \simeq S^5$ is well-known. We thus get that $$H_k(X) = \begin{cases} \mathbb{Z} & \text{if } k = 0,5,6; \\ 0 & \text{otherwise.} \end{cases}$$

From $H_6(X) \cong \mathbb{Z}$ we deduce that $X$ is orientable. It must therefore satisfy Poincaré duality, in particular $H^1(X) \cong H_5(X) \cong \mathbb{Z}$. But by the universal coefficients theorem, $H^1(X) \cong \hom_\mathbb{Z}(H_1(X), \mathbb{Z}) = 0$. This is a contradiction.

Answer 2

Call $X$ the one point compactification of $S^5\times \mathbb{R}$. Actually, I'll write $S^5\times (0,1)$ instead. Since $(0,1)$ and $\mathbb{R}$ are homeomorphic, this doesn't change the problem.

Let $\Sigma S^5$ denote the suspension of $S^5$ and let $Y$ be the quotient of $\Sigma S^5$ obtained by gluing the two cone points together.

Proposition The space $X$ is homeomorphic to $Y$.

Proof: View the suspension as $S^5\times [0,1]$ with $S^5\times \{0\}$ and $S^5\times \{1\}$ both collapsed to a point. Let $f:S^5\times (0,1)\rightarrow Y$ be the composition $$S^5\times (0,1)\rightarrow S^5\times [0,1]\rightarrow \Sigma S^5\rightarrow Y.$$

The first map has image everything except $S^5\times \{0,1\}$, but these points all get identified to a single point in $Y$. Thus, this composition provides an embedding from $S^5\times (0,1)$ to $Y$ which misses a single point. Since the closure of $S^5\times (0,1)$ in $S^5\times[0,1]$ is everything, it follows that the image of $f$ is dense in $Y$. Since $Y$ is compact, it is a one-point compactification of $Y$. However, one point compactifications are unique up to homeomorphism, so $X$ and $Y$ are homeomorphic. $\square$

So, we just need to show that $Y$ is not a topological manifold. In $\Sigma S^5$, take two small neighborhoods, one around the top cone point and another around the bottom cone point. Their projections $Y$ give two open sets in $Y$ which intersect only in a single point.

But this cannot happen in a manifold: taking a topological chart around the intersection point, we obtain a point in $\mathbb{R}^n$ with neighborhoods $U_1$ and $U_2$ intersecting at a single point $p$. By definition, there is a ball $B(p,r_i)\subseteq U_i$ for some positive radii $r_i$. But then if $r = \min\{r_1,r_2\}$, $B(p,r)\subseteq U_1\cap U_2$. That is, the intersection contains more than one point.

Thus, $Y$, and therefore $X$, cannot be a topological manifold.