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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.

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  • $\begingroup$ "...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. $\endgroup$ – autodavid Jan 8 '20 at 13:08
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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.

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  • $\begingroup$ Is there a better intuitive understanding of homotopy that you can think of? $\endgroup$ – Oliver G Jan 6 '20 at 17:41
  • $\begingroup$ Not that I'm aware of; I visualize it as the two circles shrinking to points while moving to the origin. (In general, a homotopy might involve continuous "reshaping" rather than just resizing and sliding). Is there something in particular about this visualization that you find unconvincing or unsatisfying? $\endgroup$ – Justin Barhite Jan 6 '20 at 20:26

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