I have this exercise:

If Y is path-connected, show that there is only one homotopy-class of continuous functions from $[0,1]$ to Y.

My attempt:

What I need to show is that if I have two continuous functions $f_1,f_2: [0,1]\rightarrow Y$ they are homotopic. I must find an F such that $f: I \times I\rightarrow Y$, so that F is continuous, and $F(t,0)=f_1(t), F(t,1)=f_2(t)$.

There is one way that seem very natural to construct F here, that is for every t, since Y is path-connected, there is a path from $f_1(t)$ to $f_2(t)$, this path can be written: $f_t(s): [0,1]\rightarrow Y$, where $f_t(0)=f_1(t), f_t(1)=f_2(t)$.

Then we just denote $F(t,s)=f_t(s)$.

By construction this will be a homotopy(not path-homotopy) between $f_1, f_2$ if we have that F is continuous. But how do I show that it is continuous? Or is it even continuous? What I know about continuoity is this:

$F(t,0), F(t,1)$ is continuous in t. $F(t,s)$ is continuous in s for all t. But still it is not enough, we need joint continuity. Any tips on how to show continuity?

  • $\begingroup$ I think that you need to be more discriminating when choosing the $f_t (s)$ $\endgroup$
    – basket
    Dec 24, 2015 at 9:33
  • 4
    $\begingroup$ Hint: Any function $f:[0,1]\to Y$ is homotopic to the constant function $f_0(x)=f(0)$. $\endgroup$
    – Arthur
    Dec 24, 2015 at 9:34
  • $\begingroup$ ah I see @Arthur. Shrink the loop $f_1$ down to a point and then inflate $f_2$. $\endgroup$
    – basket
    Dec 24, 2015 at 9:39
  • $\begingroup$ @basket In-between the shrink and the inflation you need to move the point to $f_2(0)$ (which is where path-connectedness of $Y$), but yes. $\endgroup$
    – Arthur
    Dec 24, 2015 at 9:41

1 Answer 1


As written, $F$ has absolutely no reason to be continuous. For example let $f_1, f_2 : [0,1] \to S^1$ ($S^1$ is the unit circle in $\mathbb{C}$) be given by $f_1(t) = 1$ and $f_2(t) = -1$, and for $s < 1/2$ I choose a path between $f_1(s)$ and $f_2(s)$ which goes through the upper half-circle and for $s \ge 1/2$ I choose a path which goes through the lower half-circle. The resulting $F$ will clearly not be continuous.

The easiest way to go as Arthur remarks in the comments) is to show that any map $f : [0,1] \to Y$ is homotopic to a constant map. Simply let $F(s,t) = f((1-s)t)$, then $F(0,t) = f(t)$ while $F(1,t) = f(0)$ is constant. Thus $f_1$ is homotopic to the constant map equal to $f_1(0)$, and $f_2$ is homotopic to the constant map equal to $f_2(0)$. (I didn't use the hypothesis that $Y$ is path-connected here.)

Then since $Y$ is path-connected, there's a path $\gamma : [0,1] \to Y$ such that $\gamma(0) = f_1(0)$ and $\gamma(1) = f_2(0)$. Let $F(s,t) = \gamma(s)$: this define a homotopy between the two constant maps $(t \mapsto f_1(0))$ and $(t \mapsto f_2(0))$. Since the relation "being homotopic" is an equivalence relation, you finally get a homotopy between $f_1$ and $f_2$: $$f_1 \sim (t \mapsto f_1(0)) \sim (t \mapsto f_2(0)) \sim f_2.$$

  • 1
    $\begingroup$ I have a doubt: what happens if $Y$ has a hole? Because you could find two path such that they are not homotopic. Do you mean free homotopy here? $\endgroup$
    – InsideOut
    Dec 24, 2015 at 9:51
  • 1
    $\begingroup$ @GianlucaFaraco Yes, it's a free homotopy. Endpoints are allowed to be moved. $\endgroup$ Dec 24, 2015 at 9:55
  • $\begingroup$ Thank for your comment! ;) $\endgroup$
    – InsideOut
    Dec 24, 2015 at 9:59
  • $\begingroup$ Thank you, the $F(t,s)=f((1-s)t)$ was very clever. $\endgroup$
    – user119615
    Dec 24, 2015 at 10:03

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