The sphere $S^2$ is not contractible I heard that in topology the sphere $S^2$ cannot be continuously deformed to a point, i.e. $S^2$ is not contractible. 
Sorry for my ignorance, but I really don't get it. Can't we just push all the points to the origin as in the following transformation?
$$f:I\times S^2\to\Bbb R^3,\quad f(t,x)=(1-t)x.$$
Then, $f(0,S^2)=S^2$ and $f(1,S^2)=\{0\}$. Isn't that a continuous deformation of $S^2$ to a point?
 A: The contraction has to take place within $S^2$. As Hans Engler says in the comments, the continuous deformation is technically a continuous map $f:[0,1]\times S^2\to S^2$, specifically, one such that $f(0,x)=x$ for all $x\in S^2$, and there is a $p\in S^2$ such that $f(1,x)=p$ for all $x\in S^2$. For each $\alpha\in[0,1]$ define $f_\alpha:S^2\to S^2:x\mapsto f(\alpha,x)$. Then $f_0$ is the identity map on $S^2$, $f_1$ is the constant map $f_1(x)=p$ for all $x\in S^2$, and each set $f_\alpha[S^2]$ for $\alpha\in[0,1]$ is a subset of $S^2$. Intuitively you can think of $f_\alpha[S^2]$ as the $\alpha$-th stage of the deformation: it’s what the deformed $S^2$ looks like at time $\alpha$, so to speak.
Your idea would work fine to deform the disk $D^2=\{x\in\Bbb R^2:\|x\|\le 1\}$ to the point at the origin: $f_\alpha[D^2]$ is simply the disk of radius $1-\alpha$, which is indeed a subset of $D^2$, the disk of radius $1$. But under your map $f_\alpha[S^2]$ is the circle of radius $1-\alpha$, which for $\alpha\ne 0$ is not a subset of $S^2$.
