To clarify the definitions here: We call a differentiable manifold contractible if the identity is homotopic to some constant function on that manifold. Furthermore we call a differentiable manifold $M$ simply connected if it is connected and every differentiable function $S^1\to M$ is homotopic to some constant function on $M$.
Now I want to show that every contractible manifold is simply connected.
My almost full proof:
Since $M$ is contractible there is a differentiable function $F:M\times I\to M$ such that $F(x,0)=x$ and $F(x,1)=p$ for all $x\in M$ with some fixed $p\in M$, where $I=[0,1]\subset\mathbb{R}$.
One easily sees that $M$ is even path connected since for every $x,y\in M$ we have that $g(t)=h^{-1}[h(F(x,t))+t(h(y)-h(p))]$ is a continuous path between $x$ and $y$, where $h$ is a suitable chart on $M$.
Furthermore for $x\in S^1$ and thus $f(x)\in M$ and $h:U\to U^{\prime}\subseteq \mathbb{R}^m$ a chart for $f(x)\in M$ and $c\in U^{\prime}$ we can define $G(x,t)=h^{-1}(h\circ f(x)(1-t)+tc)$ for every $x\in S^1$ such that $f(x)\in U$ and we get $G(x,0)=f(x)$ and $G(x,1)=h^{-1}(c)\equiv const. $ for all $x\in f^{1}(U)$. One can take a moment to check that $G$ is well defined (independent from the choice of charts, etc…) and one can even extend $G$ to $S^1$ such that $G(x,0)=f(x)$ for all $x\in S^1$ because on the possible intersections of the maps $G$ will have the same values and again it will be well defined.
My question is now how can I extend $G$ to all of $S^1$ such that the second condition $G(x,1)\equiv const.$ will be fulfilled? In other words, how can I guarantee that the aforementioned constant $h^{-1}(c)$ will be the same on every map of every chart? Is there an easier non constructive solution or does my construction not work? I guess I have to use a differential structure on $S^1$ as well.
Best Answer
I think you've gotten too bogged down with charts and what not here. Given a loop $\sigma:S^1\to M$ on $M$ and a homotopy $F$ between $M$ and a point $p\in M$, just compose the two, to get a homotopy between $\sigma$ and the constant loop. You can assume wlog that the point $p$ is the base point of the loop. Because as you note $M$ is path connected, you can consider the fundamental group at base point $p$, $\pi_1(M,p)$.