Can we find an $f$ so that the mapping cylinder $Mf$ is homeomorphic to $\mathbb{S}^3 $?

Question

I recently meet with a problem that to determine the mapping cylinder $Mf$, where $f$ is a continuous linear map from $\mathbb{T}^2 $ to $\mathbb{T}^2 $, so that $Mf$ is homeomorphic to $\mathbb{S}^3 $. I doubt if it is right. How to determine such a continuous map $f$?

Answer 1

Since this hasn't been answered: no. One can check that a mapping cylinder for a map $f:X \to Y$ is homotopy equivalent to $Y$, by retracting $X$ to $f(X)$ via $I$. In particular

$M_f$ cannot be homeomorphic to $S^3$ since they are not even homotopic (which can be seen by checking fundamental groups of $T^2$ and $S^3$.