I am trying to prove the following claim:
The following topological space, up to homeomorphism, is called the Möbius strip: $$M=[0,1]/\sim$$ where $\sim$ is the equivalence relation generated by $(0,t)\sim (1,1-t)$. Then $M$ deformation retracts to a circle.
Let $I=[0,1]$ and consider the map $F:I^2\times I\to I^2$ given by: $$F(r,s;t)=(r,tr+(1-t)s)$$ It is clear that $F$ is a deformation retract of $I^2$ to the diagonal $D=\{(r,r):r\in I\}$. It we pass to the quotient $M$ we get a continuous map $\tilde F:M\times I\to M$ given by: $$\tilde F([(r,s)],t)=[F(r,s;t)]$$
My question is:
How can I prove that $\tilde F$ is continuous?