I have a space $X$ which is defined to be the quotient of $[0,1)\times T^3$ ($T^3$ is the 3-torus) by the relation $(0,x)\sim (0,y)$ for all $x,y \in T^3$. It is a kind of cone over $T^3$, except that the interval $[0,1)$ is only half closed. I want to show that this space is not homeomorphic to $\mathbb{R}^4$.
I believe that the image of one of the $S^1$ factors in $T^3$ will be a non-null-homotopic loop, which would yield the result using fundamental groups. I can't seem to show this though, and am not sure it is a good approach.
A quick look at the cases for $T^1$ and $T^2$: in the same constructions for $T^1$ and $T^2$ I believe the fundamental groups are $0$ and $\mathbb{Z}$ respectively and in the latter case, the image of one of the $S^1$ factors is the generator. I believe this because I imagine $[0,1)\times T^2$ as a thickening of the $2$-torus with $\{0\}\times T^2$ being the outer shell which is mapped to a point in the quotient. Mapping the outer shell to a point can be thought of as first pinching the central "donut" hole, as it were, to a point. We then have something homotopic to a 3-ball but missing an interior circle, and whose boundary must be identified to a point. Identifying the boundary of a 3-ball in this way yields $S^3$ minus a circle, which is homeomorphic to $\mathbb{R}^3$ minus a line, which has fundamental group $\mathbb{Z}$ generated by a loop coming from the image of one of the $S^1$ factors of the original $T^3$. I believe the same sort of thing may happen in the case of $T^3$ but cannot currently prove it.
Any thoughts or guidance here would be greatly appreciated.
Edit: I just realized that the analysis for the $T^2$ case is wrong. When I think of identifying the boundary of a 3-ball to a point, I also have a curve in this ball each end of which is touching the boundary and which must be mapped to the same point. In this case, I am now inclined to believe that the fundamental group is also $0$ in the case of $T^2$ because that curve is homotopic to what I originally thought would be the generator.
Removing from the quotient the equivalence class of $(0,x)$ (for any $x\in T^3$) we obtain the space $(0,1)\times T^3$, which cannot be homeomorphic to $\mathbb{R}^3$ with a point removed (e.g. because the first space is not simply connected).