Let $X$ be a simply connected topological space. Let $x_0,x_1\in X$ and let $\gamma$ be a path in $X$ from $x_0$ to $x_1$. Let $g$ be a homeomorphism of $X$ with itself. Then $g\circ\gamma$ is a path in $X$ from $g(x_0)$ to $g(x_1)$.

Is $\gamma$ homotopic to $g\circ\gamma$? (NOT path homotopic, just homotopic)

I think it should be true but I can't seem to give a homotopy from $\gamma$ to $g\circ\gamma$. That is I need a continuous map $F:I\times I\rightarrow X$ such that $F(s,0)=\gamma(s)$ and $F(s,1)=g\circ\gamma(s)$ for all $s\in I$.


  • Note that since you only ask for homotopy, and not path homotopy, every map $I\to X$ is homotopic to a constant map. So it suffices when $X$ is path-connected since then the two constant paths are homotopic. – Stefan Hamcke Nov 11 '15 at 8:01
up vote 1 down vote accepted

Since $X$ is simply connected, every two paths are homotopic (almost) by definition.

  • Yes that is what I thought at first. But don't we say any two paths with the same starting point and ending point are path homotopic and not just that any two paths are homotopic? – R_D Nov 11 '15 at 7:44
  • 1
    I suppose that a simply connected space is also connected. In that case you can join the two starting and ending points with a continuous curve. Hence you assign the value of the homotopy $F$ on the boundary of $I\times I$. The function can then be extended on the interior by the definition of simply connectedness. If, instead, a simply connected space is not required to be connected, the statement is obviously false. – Emanuele Paolini Nov 11 '15 at 7:48
  • I see. You are right. A simply connected space is by definition path connected so what you have said will work. Thank you! – R_D Nov 11 '15 at 7:52

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