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).