# 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. – Sanchez Dec 27 '12 at 22:31

Perhaps I'm misunderstanding something, but I think the very definition of "contractible space" gives the answer: the identity map $\,Id.:X\to X\,$ is nullhomotopic, i.e. homotopic to some constant function $\,f:X\to\,\,,\,f(x)=x_0\,\,\,,\,\,\forall\,x\in X\,$ , and from here the "based" homotopy to be constructed is trivial.
This issue is that "contractible" doesn't require that any homotopy between your constant map and the identity is constant on $x_0$ for all $t$, which is required of based homotopies. – Jason DeVito Dec 27 '12 at 23:31
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.