Which surface is homeomorphism to mobius strip? I'm a bit confused since I read many versions of this.
First of all, I couldn't understand how to define homeomorphism. Intuitively, I think that if I can deform an object, without tearing it, to another object, then those 2 surfaces are homeomorphic to each other. 
I definitely know the Mobius strip isn't homeomorphic to the torus since torus has 2 surfaces while Mobius strip has one.
I have also heard something about that if I could "map" the Mobius strip, then I could find the surface it is homeomorphic to.
Thanks in advance.
 A: Two topological spaces $(X_1,\tau_1)$ and $(X_2, \tau_2)$ are said to be homeomorphic provided that there exists a function $f: X_1 \rightarrow X_2$ such that $f$ is one-to-one, onto, continuous, and its inverse $f^{-1}$ is continuous. By continuous, we mean that the preimage of every open set in $X_2$ is open in $X_1$. What this means is that $X_1$ and $X_2$ have the same topological structure.
Visually, especially with spaces that are formed from the subspace topology of $\mathbb{R}^n$, this means that homeomorphisms are roughly the functions that are given by "stretching without tearing, pinching, or gluing," but the rigorous definition should be used for formal proofs.
Working out whether or not two spaces are homeomorphic is in general an extremely difficult question to answer. To show that two spaces are homeomorphic amounts to spotting a homemorphism between the two, which isn't always simple. But the really hard part is often developing machinery that shows when two spaces are not homeomorphic to each other. This is typically done by finding features of spaces that are invariant under homeomorphism, and then showing that the two spaces do not have matching features. This is a good bit of the goal of algebraic topology tries to do. For instance, the topological invariant of the fundamental group offers a fast proof that the torus and mobius strip are not homeomorphic.
