In what sense is $S^n$ contractible in $S^{n+1}$?

Question

In my course of algebraic topology our professor said that although $S^n$ is not contractible, it is as a subspace of $S^{n+1}$, but he said it just as a comment and he gave the intuitive idea of why it is so but not a formal proof. However, even if I can understand that we can take the equatorial circle of the 2-sphere and slide it up through the the 2-sphere to the north pole, it doesnt't make sense to me that it is contractible when regarded as a subspace of $S^2$ but not contractible by itself. At least this doesn't make sense whith the usual definition of 'contractible space'.

So, my question is: is there a weaker notion of 'contractible space' relative to another space in which it is embedded?

My guess is that my professor meant to say that there is a homotopy $H:S^{n+1}×I\to S^{n+1}$ such that $H|_{S^{n+1}×\{0\}}=1_{S^{n+1}} $ and $H(S^n×\{1\})=x_0 \in S^n$.

Answer 1

There are two ways to interpret this, and it seems you are confused about which your professor meant.

1) As in the last sentence of your question, or

2) There is a map $G:S^n\times I\to S^{n+1}$ with $G\mid_{S^n\times\{0\}}$ the inclusion as the equator and $G\mid_{S^n\times\{1\}}$ a constant map.

Either way you interpret what your professor said is correct. For the first way, stereographically project $S^{n+1}\setminus\{\text{south pole}\}$ onto $\mathbb R^n$, apply your favorite contraction of the closed unit ball in $\mathbb R^{n+1}$, then apply the inverse of the projection. For the other interpretation, at time $t$, map the point $(x_1,\ldots,x_{n+1})\in S^n$ to $(\sqrt{1-t^2}x_1,\ldots,\sqrt{1-t^2}x_{n+1},t)$.

As for which interpretation your professor meant, you should ask them.

Answer 2

As stated in the comments, the inclusion $i_n:S^n\to S^{n+1}$ is null-homotopic. The following function is an homotopy from $i_n$ to a constant map:

$$H_n:(x,t)\in S^n\times I \mapsto (\sqrt{1-t^2}x,t)\in S^{n+1}$$

The following is an application: Let us consider the infinite ball $S^{\infty}=\cup_{n\in \mathbb{N}_0}S^n$, endowed with the final topology with respect to the inclusions $j_n:S^n\to S^\infty$, that is, a subset there is open if and only if its intersection with each $S^n$ is open. Gluing the homotopies $H_n$ (as in Hari Rau-Murthy's argument) we can get an homotopy from the identity of the infinite ball to a constant map so this space is contractible.

However, the (finite) spheres $S^n$ are not contractible. In order to prove that we have to use some invariant (under homotopy). For example they have not trivial homology.

Answer 3

As you suggest, your professor probably meant that the canonical (equator) inclusion $S^n\hookrightarrow S^{n+1}$ is nulhomotopic. Whenever a non-contractible suspace is called "contractible in" some ambient space, this is probably what is meant.

Of course, this terminology is misleading. In fact, we are frequently interested in whether a subspace $A\subset X$ is actually contractible (i.e. when given the subspace topology), e.g. for van Kampen or Mayer-Vietoris arguments.

Answer 4

The term "contractible in" doesn't exist and it was made up by your professor. We can only guess what he means. Most likely he means that some embedding $S^n\hookrightarrow S^{n+1}$ is nullhomotopic.

But this is not satisfactory at all, because all functions $S^n\to S^{n+1}$ are nullhomotopic. Also the term doesn't make sense for spaces $X,Y$ that have two different embeddings $X\hookrightarrow Y$ such that one is nullhomotopic and the other is not (I encourage you to find such examples yourself, it is not hard). So does "contractible in" apply only to $S^n$ and $S^{n+1}$? What a waste of language.

All in all, you should treat that as a very informal way of saying that some embedding (into equator?) $S^n\hookrightarrow S^{n+1}$ is nullhomotopic. I also advice asking the professor about that.