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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

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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.

The idea is also used in Lemma 1.19 of Hatcher's Algebraic Topology on page 37. Actually, this lemma is needed to show that homotopy equivalent spaces have isomorphic fundamental groups, so arguing that a point has trivial fundamental group would be a form of circular reasoning.

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