In Example 1.35, Hatcher writes in his book Algebraic Topology, the following (not paraphrased):
Let $X=S^1\times I$, and let $A$ be the quotient space obtained by defining the relation $(z, 0)\sim (e^{i\pi}z, 0)$ on $X$. Then $A$ is the Möbius band.
That is, if we identity the antipodal points on $S^1\times \{0\}$, we get the Möbius band.
I am unable to prove this. The definition of the Möbius band which I am using is the one where we identify the points $(x, 0)$ with $(1-x, 1)$ in the square $I\times I$.
I tried to "open" the cylinder by cutting along a line parallel to the axis of the cylinder. This gives us a square, and an induced equivalence relation on the square. But it doesn't look anything like the Möbius band.
Draw a fundamental polygon of your space. This is a pentagon say with sides $a,b,c,d,e$ with sides $c,e$ identified in an orientation preserving manner and $a,b$ in reverse orientation. Now cut the pentagon by the line joining the vertex where $a,b$ meet and the midpoint of segment $d$. What do you get?