I' working on this problem, and I got stuck in the middle. Could someone give me a hand? Let $X$ be a contractible space. I'd like to show that $X$ is simply connected. I already showed that $X$ is path connected. I need to show for every $x_0 \in X$, $\pi_1(X,x_0)$ is the trivial group. So, it suffices to show every loop based at $x_0$ is path homotopic to the path $e: I\to X$ defined by $e(x)=x_0$ for every $x\in X$
Let $\alpha$ be a loop based at fixed $x_0 \X$. $X$ is contractible implies there exists a homotopy $H: X \times I \to X$ such that $H(x,0)=x$ and $H(x,1)=x_0$. I tried to construct a path homotopy from $\alpha$ to $e$ using H, but I don't think mine works. I define $G(x,t)= H(\alpha(x),t)$. Then $G(x,0)=H(\alpha(x),0)=\alpha(x)$ and $G(x,1)=H(\alpha(x),1)=x_0$. But this is only a homotopy, I want $H(0,t)=H(1,t)=x_0$. I can't get it with this map.