# contractible and simply connected

Every contractible space X is simply connected because X is homotopy equivalent to a point.

Is there a direct proof of this fact? There obviously is a (free) homotopy between any loop and the trivial loop at the base point. But how to construct a based homotopy, which is required for a loop to be trivial in the fundamental group?

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Consider loop 1 formed by the starting point of your free homotopy, loop 2 formed by the ending point of the free homotopy. Then at time $t$, traverse along loop 1 up to time $t$, go through the homotopy at time $t$, and come back along loop 2. This is then a based homotopy to the trivial loop at based point. – user27126 Dec 27 '12 at 22:31

If $\varphi_t:S^1\times I\to X$ is the free homotopy between the loop $\varphi_0$ and the constant loop $\varphi_1\equiv x$, and $h$ is the path formed the images of $s_0$, i.e. $h(t)=\varphi_t(s_0)$, then define $h_t(s)=h(ts)$. At $t$ it traverses the path $h$ till the point $h(t)=\varphi_t(s_0)$, so it can be composed with $\varphi_t$, which itself can be followed by $\overline{h_t}$ to get back to $\varphi(s_0)$. So the product $h_t\cdot\varphi_t\cdot\overline{h_t}$ gives a bases homotopy between $\varphi_0$ and $h_1\cdot x\cdot\overline{h_1}$, the later being contractible.