# Deformation retract of $\mathbb{R}^3$ minus a tetrahedron

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.

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.