I am being asked to find a NON CONTRACTIBLE space $X$ such that, for any path-connected space $Y$ and any continuous maps $f,g : X \rightarrow Y$, $f$ and $g$ are homotopic.

I have thought that I definitely cannot pick $X$ to be path-connected, since otherwise picking $Y = X$, $f = Id_X$ and $g$ a constant map, I would obtain that $X$ is contractible. That said I am not too sure where to look for an actual example.

The classic example of a space that is not contractible but has trivial fundamental group, $S^2$, also cannot work here, since degree of maps is a homotopy invariant and clearly many different degree maps exist!

Any help appreciated!

  • 1
    $\begingroup$ Have you considered a connected space that is not path-connected? $\endgroup$ Feb 6, 2023 at 19:53
  • $\begingroup$ I have not. I guess the topologist's sine curve is the canonical example for this, but isn't it contractible? $\endgroup$ Feb 6, 2023 at 20:01
  • $\begingroup$ no, it's not contractible, see, e.g., here: math.stackexchange.com/questions/548681/… $\endgroup$
    – Thomas
    Feb 6, 2023 at 20:06
  • $\begingroup$ @Thomas That seems to refer to the closed topologist's sine curve or the Warsaw circle, but maybe I'm missing the spot. But isn't it clear that if we take a homotopy to the constant map (choosing the fixed point $x_0$ in the vertical line segment), the image of $\{x\}\times [0,1]$ gives a path from $x$ to $x_0$? $\endgroup$ Feb 6, 2023 at 20:15
  • $\begingroup$ Hint: try some finite spaces. $\endgroup$ Feb 6, 2023 at 20:24

1 Answer 1


You are overcomplicating things. The space you are looking for is... $\{1,2\}$ or any discrete space with at least two points. Note that a contractible space has to be connected.

As for why $\{1,2\}$ satisfies your condition I leave it as an exercise.

  • $\begingroup$ Grrrr. ......... :D $\endgroup$ Feb 6, 2023 at 20:44
  • 2
    $\begingroup$ There is an even simpler example with less than two points. $\endgroup$ Feb 6, 2023 at 21:06
  • $\begingroup$ Thank you for your help! I got it now! $\endgroup$ Feb 6, 2023 at 21:39
  • 1
    $\begingroup$ Grrrr$^2$ to @Moishe. That is really bad! But, as always with topological questions, someone should be the custodian of the empty set. $\endgroup$ Feb 6, 2023 at 22:27
  • $\begingroup$ Just to check I am right: if I have a function $f$ from {1,2} to Y, would the correct homotopy be $H: {1,2} \times I \rightarrow Y$ which is defined by $H(1, -) = f(1)$ and $H(2, t) = p(t)$ where $p$ is the path starting at $f(2)$ and ending at $f(1)$? This would give me homotopy between f and the constant map at f(1), and then since all constant maps are homotopic in a path connected space, I obtain homotopy between f and any other map. $\endgroup$ Feb 6, 2023 at 22:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .