Can the (extended) Möbius strip be found as the torus $T^2$ minus some embedded $S^1$?

Question

Denote $M = \text{{extended Möbius strip}} = \text{{mobius strip with deleted boundary}}$.

I can recover a cylinder $S^1 \times \mathbb{R}$ if I setwise-remove either of the two meridians of the torus.

Let $\phi: [0,1] \times [0,1] \rightarrow T^2$ denote the canonical quotient map. Then I suspect the diagonal $\psi: [0,1] \rightarrow [0,1] \times [0,1]$ given by $\psi(x) = (x,x)$ has the property that $T^2 \backslash \text{im}(\phi \circ \psi) \cong M$. Can anyone confirm this?

Answer 1

This is not true, any open subset of the torus is an orientable manifold, but $M$ has no orientation. Most books on topology will cover this.