How to prove a Möbius band is not equivalent to a band A band (or a disk with a hole in it) can be created by gluing two edges of a square in the same direction, while a Möbius band can be created by gluing two edges of a square in opposite directions. These two spaces are not homeomorphic (right?), but proving they are different seems much more difficult. They are homotopy equivalent, both connected -- in general they seem to share the obvious properties.
What is the method of proving these two spaces are different?
I've worked my way through a good chunk of point-set topology but I don't have any foundations in algebraic topology. Does this problem require heavy algebraic topology to solve, or can I do it with point set topology?
 A: Here is a result with a bit of point set.
Point set fact: if $X,Y$ are hausdorff and homeomorphic, then they have homeomorphic one point compactifications.
With this, you can show that the one point compactification of the mobius band is hoeomorphic to $\mathbb RP^2$. This can be seen explicitly by considering the punctured projective plane.
The one point compactification of the cyllinder is a sphere with two points identified.
You can either stop here if you are satisfied with orientability (or embeddings into $\mathbb R^3$.)
Otherwise, note that the latter space is homotopic to $S^2  \vee S^1$.
Now, you have the job of distinguishing $S^2 \vee S^1$ and $\mathbb RP^2$ up to homotopy is not so bad. For example, what are the possible maps $\pi_1(\mathbb RP^2) \to \pi_1(S^2 \vee S^1)$? Or said differently, there is a loop of order $2$ in $\mathbb RP^2$, is there anything like that in $S^2 \vee S^1$? So on, and so forth.
A: Let's talk about the "edge" of a strip rather than its boundary. I hope everyone
agrees that the edge of the untwisted strip consists of two circles, and that
of the Mobius strip consists of one circle, so if we can characterise the edge
topologically, it will show these two spaces are not homeomorphic.
Let $X$ be one of the two spaces, and let $x$ be a edge point. Then $x$ has
a neighbourhood basis consisting of sets $U$ having the property that $U\setminus\{x\}$ is simply connected. Think of a family of "D-shaped" subsets shrinking
towards $x$. But if $x$ is a non-edge points, then this is not possible. Each neighbourhood basis of $x$ contains sets homeomorphic to open subsets of $\Bbb R^2$. Deleting a point from such a set will leave a non-simply connected
set (consider a circle centred at the deleted point).
