Middle of mobius band is a circle We define the Möbius band $M$ as the space $I \times I$ when identifying $(0,t)\sim (1,1-t)$ through $\rho: I \times I \to M$. We wish to prove $S=\rho(I \times \{ \frac{1}{2}\})$ is a circle. The solution was to define a function
$f:I \times \left\{\frac12\right\} \to S^1$ by 
$f\left(t,\frac12\right)=(\cos(2\pi t), \sin(2 \pi t))$ and note it is continuous and closed (since it's from a $T_2$ space to a Compact space). How does it help? I assume we try to get a homeomorphism $S \to S^1$ somehow.
 A: Use the universal property of quotient maps. Please have a look at the following diagram:

Here $p$ is the quotient map to the Möbius band and $q$ is the restriction of $p$ on the middle line. Now there are two ways to topologize the set $p\left(I\times\left\{\frac12\right\}\right)$: We can regard it as a subspace of $M$, or we give it the final topology with respect to $q$ which means we see at as a quotient of $I\times\left\{\frac12\right\}$. There are cases where these topologies are different, and we see that they are equal iff $q$ is a quotient map. There are criteria which enable us to decide if this is the case. For example, if $A$ is a closed and saturated subspace of $X$, then the restriction of a quotient map to $A$ is again a quotient map. As $I\times\left\{\frac12\right\}$ is indeed closed in $I×I$ and saturated with respect to $p$, we conclude that $q$ is a quotient map.
Now $f$ as you defined it was continuous, surjective, and was compatible with the relation (meaning that $f(x)=f(y)$ whenever $q(x)=q(y)$). By the universal property of quotient maps it thus factors as $\tilde f\circ q$, where $\tilde f$ is uniquely determined and surjective.
Recall what we need for $\tilde f$ to be injective and open.
