# Homotopy equivalence of the space obtained by removing a point from 3-torus

Consider the 3-torus $$S^1\times S^1\times S^1$$ represented by the following fundamental domain Now remove one point from this space. Then to which well-known space will it be homotopic equivalent?

My guess is that it will be homotopic equivalent to $$S^1 \vee S^2$$ since the space obtained by removing one point from 2-torus $$S^1\times S^1$$ is homotopic equivalent to $$S^1\vee S^1$$ (source). • Just a naive question: what is your intuition leading to $S^1\vee S^2$ rather than $S^1\vee S^1 \vee S^1$ or $S^1 \vee RP^2$ for example? Apr 13, 2021 at 7:56
• Not sure what it is, but van Kampen tells you that you'll get a fundamental group of $\mathbb{Z}^3$ Apr 13, 2021 at 7:59
• @Didier I am imagining that if we remove a point from one of the faces of the cube then that part should deform to $S^1$ and the rest of it should deform to $S^2$ since if we would have removed 2-skeleton from this CW complex then would have obtained a space homeomorphic to $S^3$. I have no good logic to support my guess. Apr 13, 2021 at 8:10
• I don't think it will be $S^1\vee S^2$. Removing a point should correspond (up to homotopy equivalence) to removing the top cell of its CW-structure. Therefore, it should be $S^1\vee S^1 \vee S^1$ with three $2$-cells attached via the words $aba^{-1}b^{-1}$, $aca^{-1}c^{1}$, and $bcb^{-1}c^{-1}$ where $a$, $b$, and $c$ denotes the three $1$-cells. I don't know which well-known space this is homotopy equivalent to. Apr 13, 2021 at 8:19
• I'm not sure to which well-known space is it homotopic equivalent to, but it's like three $T^2$'s glued together along their meridian and longitudes. I don't think it's possible to simplify this anymore. Apr 13, 2021 at 11:26

There is a CW-structure for $$S^1\times S^1\times S^1$$ with a single $$0$$-cell, three $$1$$-cells, three $$2$$-cells, and a single $$3$$-cell. The picture of this is the one from the question. For more details, see this post. We then get that the $$1$$-skeleton is $$S^1\vee S^1\vee S^1$$. The $$2$$-cells are attached via the words $$aba^{-1}b^{-1}$$, $$aca^{-1}c^{-1}$$, and $$bcb^{-1}c^{-1}$$ where $$a$$, $$b$$, and $$c$$ denotes the three $$1$$-cells. The $$2$$-skeleton can also be seen as hollowing out the cube from the question. Note that removing a point from $$S^1\times S^1\times S^1$$ results in a space that deformation retracts to the $$2$$-skeleton. This is analogous to the case of the $$2$$-torus which you have drawn. Let us denote this space with $$X$$.
Now let me give another geometric description of $$X$$. Let us attach the $$2$$-cells one at a time. Attaching the $$2$$-cell corresponding to the word $$aba^{-1}b^{-1}$$ yields the space $$S^1\vee (S^1\times S^1)$$ as seen in the picture: Then one glues another $$2$$-cell via the word $$aca^{-1}c^{-1}$$. This yields a wedge of two tori with two $$1$$-cells from their $$1$$-skeleta identified. Attaching the last $$2$$-cell yields a wedge of three tori such that each torus has the two $$1$$-cells identified with a $$1$$-cell from each of the other tori respectively. This yields another description of $$X$$. Let $$(S^1\times S^1)_i$$ for $$i=1,2,3$$ be three tori. Then the space can be described as the quotient $$X=\frac{(S^1\times S^1)_1\vee (S^1\times S^1)_2 \vee (S^1\times S^1)_3}{(z,*)_1\sim (z,*)_2,\, (*,z)_1\sim(*,z)_3,\, (*,z)_2\sim(z,*)_3}$$ where $$*$$ denotes the basespoint of the circle. (For example pick $$*=(1,0)\in S^1$$).