The fact that $H_n(S^n,S^n-\{x\})\cong\mathbb{Z}$ is a bit unintuitive to me, can anyone give me some insight?

Question

The fact that $H_n(S^n,S^n-\{x\})\cong\mathbb{Z}$ is a bit unintuitive to me, can anyone give me some insight?

This is a common fact in any introduction to algebraic topology course, it is achieved by using an exact sequence of pairs and then showing that $H_n(S^n,S^n-\{x\})\cong H_n(S^n)$.

It seems strange to me though.. When I think about $H_n(S^n,S^n-\{x\})$ as the homology group of $S^n$ except removing $S^n - \{x\}$, it seems that only one point should be left, and so the $nth$ homology group should be trivial. I realize that $H_n(S^n,S^n-\{x\})$ is not a "good pair" as Hatcher would say, and that my way of thinking was definitely oversimplifying things, but I was wondering if anyone could shed some light in general onto this for me. Thanks!

Answer 1

I like to think of $H_n(X,A)$ as "trying" to be $H_n(X/A)$. If $(X,A)$ is a "good" pair this is a literal equality, but in general it's more of a heuristic.

If you try to compute $H_n(S^n/(S^n - \{x\}))$, you run into the fact that $S^n/(S^n - \{x\})$ is a two-point space, and turns out to be contractible. My intuition for why the homologies fail to be equal is that $S^n - \{x\}$ "surrounds" $x$ in a way that can't be captured when we quotient it away, because we simply don't have enough points or open sets. The "right" way to fix this is to replace $X/A$ with the cofiber of the inclusion (essentially a "homotopically good version of the quotient"), but I don't think this is very intuitive if you're learning from Hatcher's perspective, so here's a heuristic that won't always work but does give some insight:

It's a general fact that homotopy equivalences preserve homology. In our case, we can contract $S^n - \{x\}$ to $S^n - X$, where $X$ is a little (closed) circular disk around $x.$ The pair $(S^n, S^n - \{x\})$ is then homotopy equivalent to the pair $(S^n, S^n - X)$, so in particular they have the same homology. $(S^n, S^n - X)$ is not a good pair, but intuitively $H_n(S^n, S^n - X)$ "wants to be" $H_n(S^n/(S^n - X)).$ But $S^n / (S^n - X)$ just looks like $S^n$, so its $n$th homology group is $\mathbb{Z}.$

You need to use some sort of excision argument like in the other answers to get a rigorous proof, but in general this heuristic of "replace spaces with homotopy equivalent spaces with a little bit more wiggle room" is a good way of making your intuitive guesses line up with the actual answer.

Answer 2

I don't necessarily know what will be intuitive to you, but here's an idea to understand relative homology : a cycle in $ C_n(X,A) $ is a chain in $ X $ whose boundary lands in $ A $.

Thus a cycle $ y $ that isn't a boundary is a chain in $ X $ whose boundary lands in $ A $ and that isn't itself a boundary + something that lands in $ A $.

In $ X= S^n, A=S^n \setminus \{x \} $ you notice that if the boundary of $ y $ lands in $ S^n \setminus \{x \} \simeq \mathbb{R^n} $ then this boundary is a cycle in $ \mathbb R^n $, but this has vanishing homology, thus it must also be a boundary in $ \mathbb R^n \simeq S^n \setminus \{x \} $ and thus in $ S^n $, thus it becomes zero in homology.

Thus the only relative homology is already absolute homology of $ S^n $

(I don't know if that counts as intuition because all I did was basically run the long exact sequence argument in this special case )

Answer 3

Since $S^n$ is a manifold, we can pick a chart and use excision to get an isomorphism

$$ H_n(S^n,S^n - x)\cong H_n(\mathbb{R}^n,\mathbb{R}^n - 0) $$

This helps us to remember that this relative group is only concerned with local behaviour at the point $x$. (In fact we can make this simplification for any manifold with basepoint $(M,x)$.) One way to show this second group is $\mathbb{Z}$ is to use homotopy invariance and excision again to get

$$ H_n(\mathbb{R}^n,\mathbb{R}^n - 0) \cong H_n(\mathbb{R}^n,\mathbb{R}^n - int(D^n)) \cong H^n(D^n,S^{n-1}) \cong \tilde{H}_n(S^n) \cong \mathbb{Z}$$

In fact we define a homology $n$-manifold to be a topological space $X$ such that $H_n(X, X-x)\cong \mathbb{Z}$ for every $x\in X$. By this argument every manifold is a homology manifold.

We can also produce explicit elements of this group. As Max points out in his answer, the relative cycles are the singular simplices $\sigma\colon \Delta^n \to \mathbb{R}^n$ whose boundaries do not contain $0$. We can construct a function from $\Delta^n = convex\ hull(e_1,\dots, e_{n+1}) \subset\mathbb{R}^{n+1}$ to another real vector space by choosing images of each basis vector $e_i$ and then extending by linearity. In particular we can define $\sigma$ by $\sigma(e_i) = e_i$ when $i\leq n$, and set $\sigma(e_{n+1})=(-1,\dots, -1)$. Now $\sigma$ is actually an embedding, and $\sigma(\Delta^n)$ contains $0$ in its interior so is a relative cocycle. There is a homotopy equivalence of pairs $(D^n,0)\simeq (\sigma(\Delta^n),0)$, so if we use $\sigma(\Delta^n)$ in the above sequence of isomorphisms then we would see $\sigma$ gets sent to a generator of $H_n(S^n)$; therefore $\sigma$ also represents a generator of $H_n(\mathbb{R}^n, \mathbb{R}^n-0)$. To get a singular simplex representing "$-\sigma$", precompose $\sigma$ with any odd permutation of the basis vectors $e_1,\dots,e_{n+1}$ (in particular you could just switch $e_1$ and $e_2$).