Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been looking over problems in algebraic topology and figuring out questions dealing with homotopy equivalent spaces. I noticed the following problem, but can't formally verify my answer.

Determine whether or not $X = \mathbb{R}^3 \setminus \{p\}$ and $Y = \mathbb{R}^3 \setminus l$ are homotopy equivalent, where $\{p\}$ is a point and $l$ is a line.

Intuitively, it seems as though they are not, but as mentioned, I'm having a difficult time showing this. Would I be able to somehow show they are not using their fundamental groups? So far, I know that $\pi_1(X) = \{e\}$.

Can anyone help?

Thank you!

share|improve this question

2 Answers 2

up vote 5 down vote accepted

I not sure what it means to directly show 2 spaces are not homotopy equivalent - using fundamental groups is a fine way to do it.

To that end, the space $Y$ deformation retratcs onto $\mathbb{R}^2 - \{p\}$ where $p$ is a point on $l$. This space further deformation retracts onto $S^1$. What's $\pi_1(S^1)$?

share|improve this answer
I edited that part out of the question. I know that $\pi_1(S^1) = \mathbb{Z}$. –  josh Jun 2 '12 at 23:51
@josh: Exactly - so what does that tell you about $X$ and $Y$. –  Jason DeVito Jun 3 '12 at 0:20
Would this imply that they are not homotopy equivalent? Is there a result saying that non-isomorphic fundamental groups implies the spaces are not homotopy equivalent? –  josh Jun 3 '12 at 0:30
I was trying to make sense of the deformation rectraction of $Y$. Is my reasoning correct? $Y$ deformation retracts onto $S^2 - \{p, q\}$, where $p$ and $q$ are points on $l$? Then, $S^2 - \{p, q\}$ deformation retracts onto $\mathbb{R}^{2} - {p}$. –  josh Jun 3 '12 at 0:44
@josh: Homotopy equivalent spaces have isomorphic fundamental groups. If one space deformation retracts onto another, then both have the same fundamental group. Hence, yes, this shows $X$ and $Y$ are not homotopy equivalent. For your second question, yes, it does deformation retract onto $S^2 -\{p,q\}$, but that's not how I was thinking of it to myself. For definiteness, take $l$ to be the $z$ axis. Then the map sending $(x,y,z)->(x,y,tz)$ is the identity when $t = 1$ and maps $\mathbb{R}^3-l$ onto $\mathbb{R}^2-\{(0,0\}$ when $t = 0$. –  Jason DeVito Jun 3 '12 at 0:48

In algebraic topology one has topological invariants, i.e. quantities that are invariant under homotopy equivalence. If you can calculate such a quantity for each of your spaces and the output is different then you are safe to conclude that the spaces are not homotopy equivalent. Examples of such invariants are homotopy groups, homology groups (these two are related but the latter is easier to compute). And if the homology groups turn out to be equal you could try to compute their cohomology ring. E.g. take a look at $\mathbb{S}^2\vee\mathbb{S}^4$ and $\mathbb{C}P^2$; their homology groups coincide but they have different ring structure.

I suggest you try to compute these quantities for yourself. It is good practice to roll up your sleeves and get your hands dirty on some elementary spaces!

edit: If you did not already know: the fundamental group (first homotopy group) is one such topological invariant. This should leave you fit to answer the question. I hope that helps.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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