How is that $S^1$ is not contractible? It is stated in Wikipedia (and other pages too) that the spheres $S^n$ are all not contractible. 
Take $n=1$. Would anyone explain to me why $$S^1\times [0,1]\to S^1$$$$(e^{2\pi i t},s)\mapsto e^{2\pi i ts}$$is not an homotopy between the identity and a point?
 A: $$e^{2\pi i}=1,$$
so that
$$(e^{2\pi i},s)=(1,s)$$
but
$$(e^{2\pi i},s)$$
 is mapped to $e^{2\pi is},$ while $$(1,s)$$ (which corresponds to $t=0$) is mapped to $1.$
For $0<s<1$ you have
$$ e^{2\pi is}\not=1$$
which shows that your map is not well-defined on the circle. (Not to talk about continuity.)
A: We can easily prove that S¹ in not contractible.
Let's recall some definations and results for the proof.
★Simply Connected Space: A path Connected space X is called as simply connected, if every closed curve in X is a null homotopy.
After studying Fundamental groups, one can define the term as: "A path Connected space X is called Simply Connected, if it's fundamental group is trivial.
★RESULT: A contractible space is Simply Connected.
Now we are all set to prove that, S¹ is not contractible.
As we stated above,A contractible space X is Simply Connected, i.e. fundamental group defined on contractible space X at any point x is trivial.
But we know that fundamental group π1(S¹,z), for any z in S¹, is Homeomorphic to Z(the set of all integers) which can't be trivial.
Hence S¹ in not Simply Connected, implies S¹ is not contractible.
