I have a fairly simple problem with understanding the definition of contractible spaces, but just can't solve it.
Clearly, the unit disk $D^2$ in $\mathbb{R}^2$ is an example of a contractible space, while $S^1$ is not. This means that the identity map in $D^2$ is homotopic to a constant map, which doesn't happen in $S^1$. The constant map, however, is always continuous, so I don't see how there wouldn't exist a homotopy $H: I\times S^1\rightarrow \{ p\} $ between the identity and the constant map. Why isn't it the case, and why is it possible for $D^2$?
Why, for example, is the function $H: I\times S^1\rightarrow \{ p\} $ that maps $x\in S^1$ to $p$ by "walking $x$ in the circle until it gets to $p$" not a homotopy?
Best Answer
Intuitively, for $S^1$, the identity cannot be deformed to a constant without leaving $S^1$. The homotopy, if it exists, gives a deformation of the space to a point. But there's a hole in $S^1$. Not the case for $D^2$.
Look at the definition of homotopy to see why: $H:S^1\times I\to S^1$. You have to be able to stay within the space while deforming it. There's no way to pull $S^1$ apart and deform it to a point in $S^1$ continuously.