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Question from a past exam in an introductory course in algebraic topology.

Call $X$ the space $\mathbb{R}^3$ minus the edges (one dimensional) of a tetrahedron. Does $X$ deformation retract onto a CW complex with two 0-cells and four 1-cells attached between those two points?

My answer would be no, because I found a deformation retraction of $X$ onto a sphere together with its center and 4 rays, but I'm not quite sure of it. But if this the case, then the answer to the original question would be no because a deformation retraction induces a homotopy equivalence and this is not possible since the latter space has e.g. non trivial second homology group.

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This question seems to have been specifically designed to make you realise that the homology groups of a space can sometimes pick out homotopical information about a space which is not contained in the fundamental group alone. In this case, we have $X$ which is homotopy equivalent to $S^2\vee S^1\vee S^1\vee S^1$ (as you rightly pointed out, although were one small step away from reaching) and then we have a space which I'll call $Y$ which is homotopy equivalent to $S^1\vee S^1\vee S^1$.

Now then, $X$ and $Y$ are both connected and path connected (the first place we'd look to see if two spaces were homotopy equivalent). Also, $X$ and $Y$ have the same fundamental group by a trivial application of Van-Kampen's theorem giving $$\pi_1(X)\cong\pi_1(Y)\cong F_3$$ where $F_3$ is the free group on three generators. This also tells us that $H_0$ and $H_1$ will be isomorphic as $H_0$ counts path components and $H_1\cong\pi_1^{ab}$ the abelianisation of the fundamental group.

Where is the next place to look then? There are a few approaches which I'd suggest initially. The first would be to note that $\pi_2(X)$ is non-trivial (generated by the inclusion of $S^2$ in to the wedge product) and $\pi_2(Y)$ is trivial as $Y$ is a graph. However, you may not have met the higher homotopy groups yet so disregard this approach if you haven't.

Next, I would suggest just calculating homology groups. This is rather easy if you've had enough practice calculating simplicial homology and you should find that $H_2(X)$ is non-trivial (again generated by the inclusion map) whereas $H_2(Y)$ is trivial (using a standard dimension argument). This is probably the approach your text/teacher expects.

The last approach I would suggest, which you may be comfortable with, is to find the universal cover of $X$ and $Y$. We know that $Y$ is a graph and so its universal cover $\tilde{Y}$ is contractible, however $X$ has a universal cover $\tilde{X}$ which is not contractible (essentially by the same argument that $S^2$ is not contractible, the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map). This last approach is probably more contrived than the other two approaches though as the easiest way to show that the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map is to use cellular homology.

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  • $\begingroup$ Thank you very much for your clear answer! $\endgroup$ – random Jun 23 '13 at 21:17

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