Homotopy intuition about why two disjoint disks in $D^2$ are homotopic to a point in $D^2$

83 Views Asked by At

If $Y$ is contractible and $X$ is any space, then any $f,g : X \rightarrow Y$ are homotopic.

Let $Y=D^2$ the unit disk in $\Bbb R^2$ and let $X$ be two smaller disjoint disks lying within $D^2$.

Let $f : X \rightarrow Y$ by the inclusion map of the two disks into $D^2$ and let $g : X \rightarrow Y$ be the map sending all points to the $0$ point.

Since the two maps are homotopic, the two disks in $D^2$ are homotpic to a point.

My questions are:

(1) What exactly is a homotopy that continuously deforms two disks into a single point?

(2) Is there a better "intuitive" understanding of homotopy besides: shrinking and expanding objects by compressing and identifying "touching" points together or vice versa?

Because the fact that two disks are homotopic to a point in $D^2$ defies the visual analogy I created when thinking about homotopy.

1

There are 1 best solutions below

2
On

Since $Y = D^2$ is not just contractible, but in fact a convex subset of $\Bbb R^2$, a straight-line homotopy works: $$ H : X \times I \to Y, \qquad (x, t) \mapsto (1 - t)x. $$ This homotopy shrinks the two circles down to points while simultaneously moving them along a straight line to the origin.