Is $\textrm{im}(f)$ homeomorphic to the torus less the inner equator?

Question

Consider the map $id_{S^1}\times f:S^1\times [0, 1]\longrightarrow S^1\times S^1$ where $f:[0, 1]\longrightarrow S^1$ is given by $$f(t)=(\cos(\pi t), \sin(\pi t)).$$ Is it true that $\textrm{im}(f)$ is homeomorphic to the torus $S^1\times S^1$ less the inner equator:

In the above picture the blue part is the equator to be removed.

Answer 1

Not quite. Think about the fact that the equator is a closed subset of the torus. What does that say about its complement (open vs closed)? How does that compare to $\operatorname{im}(f)$?