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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!

2 Answers 2

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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 $\pi_1(S^1)$?

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    I was unaware of that useful fact. I like your method of thinking for that deformation retraction. Thank you!2012-06-03
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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.