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

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.

• "...defies the visual analogy I created when thinking about homotopy" Do you mean homotopy equivalence? If so, begin homotopic to inclusion does not mean homotopy equivalence. – autodavid Jan 8 '20 at 13:08

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.