Rotman: Examples of identifications that are not homeomorphisms I'm working through Rotman's Algebraic Topology, and he defines an identification as a map this is continuous and an open mapping, i.e. $f: X \to Y$ is an identification if $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. He then goes on to prove corollary 1.10 that gives a way of constructing a homeomorphism from an identification.
Identifications seem like such nice maps to me that I can't think of a good example of one that is not a homeomorphism. Having such an example would help make this corollary more concrete.
 A: Note that what Rotman is calling an identification is often called a quotient map, which are maps that we typically think of as "identifying" points of some space to get a new space. Other examples of quotient maps that are not homeomorphisms are 


*

*The quotient map $\Bbb R^{n+1}\smallsetminus\{0\}\to\Bbb R\Bbb P^n$, where $\Bbb R\Bbb P^n$ is real projective $n$-space. This isn't a homeomorphism because the projective space is actually compact, while the punctured Euclidean space is not.

*The quotient map that identifies corresponding points in two copies of $\Bbb R$ except for their origins. The resulting space is known as the line with two origins and it isn't Hausdorff, though the domain is Hausdorff.

*The quotient space where we identify two points $\alpha,\beta\in\Bbb R$ if and only if $\alpha-\beta$ is an integer multiple of $2\pi$ is homeomorphic to the circle, but it isn't homeomorphic to $\Bbb R$ because again the circle is compact while $\Bbb R$ is not.

*The quotient map that identifies all points in a given space. The resulting space is a singleton, which is seldom homeomorphic to the original space.
A: Never mind, by thinking a bit more carefully, I have an idea: $I \times I$ is the unit square, $h$ is an identification that takes the square to a cylinder $C$ by mapping both the top and bottom edges to the same part of the cylinder, which causes $h$ to not be a bijection. By modding these two edges together in the quotient space we then get a homeomorphism from $I \times I \: / \sim \to C$ simply because we made $h$ bijective.
