Contractible spaces has trivial fundamental group.

I have to prove the following: Show that if $X$ is contractible (the def. I have is that $I:X\rightarrow X$ the identity function is homotopic to the constant function $p$ for some $p\in X$), then its fundamental group is trivial.

Here is my attempt: Since the space is contractible, it is path connected (every point will be contracted to a fixed point, then for any two arbitrary, take the path to the fixed point, and then walk backwards on the other path, this gives you a path between any two points), I can assume that $\pi_1(X,p)$ is such that $p$ is the contraction point (since in a path connected space, all the groups obtained at different origins are isomorphic), then, let $h$ be my homotopy between the identity and the constant map, so we have: $$h(0,-)=I(-)$$ $$h(1,-)=p$$were $I$ is the identity on $X$ and $p$ is the constant path. Let $\gamma\in \pi(X,p)$. I want to show $[\gamma]=[p]$, so I thought of this homotopy, $F:[0,1]^2\rightarrow X$ via: $$F(t,s)=h(t,\gamma(s))$$Then, $F(0,s)=h(0,\gamma(s))=I(\gamma(s))=\gamma(s)$, $F(1,s)=p$, and lastly $F(t,0)=F(t,1)=h(t,\gamma(0))=h(t,p)$. I do not know how to argue why $h(t,p)=p$ for a contraction to $p$. It might be that my $F$ is incorrect too. (Note that $F$ is clearly cts, since $h$ is and $\gamma$ is as well).

Your idea is correct, but the moving $p$ is detail to be taken care of. I like to fix bad homotopies by messing with the square, since the space is too complicated. Imagine starting at the bottom-right corner of the square. Then going backwards along the bottom is going along the path $\sigma(t)$ of the point $p$. That is, $\sigma(t) = h(t,p)$. If we then go up, this goes along our path $\gamma$, and finally we go along the top, tracing $\sigma$, in the forward direction. So your square seems to be a "weird homotopy" from $\sigma^{-1} \ast \gamma \ast \sigma$ to the constant path, with the former path along three of the sides instead of just one. However, I claim that a clever transformation $I^2 \rightarrow I^2$ can turn this "weird homotopy" you defined into an honest homotopy from $\sigma^{-1} \ast \gamma \ast \sigma$ to the constant path.
EDIT: You must, of course, then figure out how to move from this to the fact that $\gamma$ is homotopic to the constant path. For this, you will either apply a nice theorem or attempt another fudge map for your square.
• Zach L: I'd be interested in hearing what your "clever transformation" $I^2 \to I^2$ is. – Jesse Madnick Jul 27 '15 at 8:15
If your space X is contractible by hypothesis then you have an homotopy equivalence between X and $\lbrace x_0 \rbrace$ (the space with only one point). Then it is not difficult to prove that the homotopy equivalence induce an isomorphism between the 2 fundamental groups.