A bijective map that is not a homeomorphism Let the map $f:[0,2\pi)\to S^1;  t\mapsto e^{it}$. This map is clearly continuous and bijective. The inverse map associates to each point $z$ on the circle, the argument of
$z$ modulo $2\pi$. it seems like $f$ is an homeomorphism, i know it is not but i don't see why? Is there any topological invariant that indicates that $[0,2\pi)$ and $S^1$ are not homeomorphic ?
 A: One way to see why the inverse mapping cannot be continuous is to note that $S^1$ is compact while $[0,2\pi)$ is not.
A: Consider the sequence $z_n:=e^{i(-1)^n/n}$ in $S^1$.
Then $(z_n)$ converges to $e^{0}=1$.
But if you apply the inverse map $f^{-1}$, you obtain the sequence $(x_n)$ given by $x_{2n}=1/(2n)$ and $x_{2n+1}=2\pi-1/(2n+1)$.
Since the latter does not converge, the inverse map $f^{-1}$ is not sequentially continuous. Hence it is not continuous.
Of course, this elementary answer does not highlight the topological invariants you are looking for. 
A: The inverse map is not continuous. 
A: Homeomorphisms preserve the structure of topologies...i.e. the open sets.
Continuous maps preserve the structure one way since a function of topological spaces is continuous if and only if the preimage of open sets is open.
However (bijective) homeomorphisms do this AND preserve open sets the other way, i.e. the image of open sets is now required to be open too.
Anyway, if you draw some pictures you will convince yourself that the open sets are not preserved.
A: Lots of answers, but this should also be mentioned: Look at the space $[0,2\pi)$ and ask what the open neighborhoods of $0$ are in that space.  They are of the form $[0,\varepsilon)$.  If every open neighborhood of $0$ were to reach into both that left end of the interval and also into the right end, looking like $[0,\varepsilon_1)\cup (1-\varepsilon_2,1)$, then $[0,2\pi)$ would be homeomorphic to the circle.
