Mobius strip homotopically equivalent to filled out torus

Question

Why is the Mobius strip homotopically equivalent to $S^1 \times D^2$? The latter is a filled out torus. How do we deform it to get the Mobius strip? Im not very familiar with homotopic equivalence, so Im looking for intuition.

Answer 1

One can think of the Mobius strip as the quotient space

$$[0,1] \times [0,1] /(0, x) \sim (1, 1-x)$$

So there is an $\mathbb S^1$ in the middle: $[0,1] \times \{1/2\}$.

Answer 2

To elaborate on John's answer, the quotient of the function $(x,y,t)\mapsto (x,y(1-t)+t/2)$, $(x,y,t)\in [0,1]\times [0,1]\times [0,1]$ is a deformation retraction from the Mobius strip onto a circle.