Are these quotient spaces homeomorphic to a cylinder and to the Möbius Strip?

Question

Consider for $[0,1]\times [0,2]$ the function $f:\{0\}\times [0,1]\to \{1\}\times [0,1]$ given by $f(0,x)=(1,x)$. Prove that the quotient space given by this $f$ is homeomorphic to the cylinder and do the same for $f(0,x)=(x,1-x)$ (this last quotient space is called Möbius Strip).

Any idea of how to start this problem?. Usually I like to give a sketch of what I tried, but this one let me baffled, I don't even have an idea to start with.

What I do have is a few questions about it:

(1) Is $f$ the equivalence relation?, then the quotient space here is $([0,1]\times [0,2])/f$?

(2) What is a cylinder?. Asking aside from the classical definition, how is a cylinder defined in topology?.

(3) $f$ seems to be bijective, how can a bijection functions define equivalent classes in order to have a quotient space?

Answer 1

1. Here I think the implied relation is $x \sim y \iff f(x) = y$ or $f(y) = x$ (we should always have that $x \sim x$ too).
2. I don't know that there is a particularly topological definition of the cylinder, if you use something like $S^1 \times [0,1]$ (where $S^1$ is the unit circle) that should be fine. You could alternatively find an explicit expression for the cylinder as a subset of $\mathbb{R}^3$.

I'm not really sure what the last part of the question is asking as it seems to define the Mobius strip as the quotient space there, so unless you have seen another expression for the strip you couldn't prove it is homeomorphic to it. I guess they just want you to show the equivalence relation works out ok!