I'm reading Hatcher's book Algebraic Topology and on page 12 he writes:
… It is less trivial to show that there are path-connected spaces that do not deformation retract onto a point…
The definition of deformation retract is given as:
A deformation retraction of a space $X$ onto a subspace $A$ is a family of maps $f_t :X→X$, $t \in I$, such that $f_0 = 1 $ (the identity map), $f_1(X) = A$, and $f_t |A = 1$ $\forall t$. The family $f_t$ should be continuous in the sense that the associated map $X×I→X, (x,t) \rightarrow f_t(x)$ is continuous.
Now my question is: how is it possible for a space $X$ to be path-connected but not retractable to a single point? Can one not just pick $x \in X$ and then retract any $y \in X$ along the path between $x$ and $y$?
Best Answer
Suppose $X$ is path-connected, then for any $y \in X$, and for any $x$, there is a path $\phi_x : I \to X$ such that $\phi_x(0)=x$ and $\phi_x(1) = y$.
With those maps, you can form a map $\phi : X \times I \to X$, with $\phi(x,t) = \phi_x(t)$. This map has all the properties you would want of a deformatino retract from $X$ onto $y$, except that it has no reason at all to be continuous.
An example of a non retractable path connected space is the circle.
Let's say you pick $y = (1,0)$, and decide to connect any $x$ to $y$ in a clockwise motion, via the top of the circle. Then the map you $\phi$ you will obtain will not be continuous at $(y,t)$ for any $t \in ]0;1[$, since the neighboor points below $y$ will travel all around the circle, while $y$ stays where he is.
And as Hatcher says, it is not trivial to show that the circle is not retractable onto a point.