I' working on this problem, and I got stuck in the middle. Could someone give me a hand? Let $\displaystyle X$ be a contractible space. I'd like to show that $\displaystyle X$ is simply connected. I already showed that $\displaystyle X$ is path connected. I need to show for every $\displaystyle x_0 \in X$, $\displaystyle \pi_1(X,x_0)$ is the trivial group. So, it suffices to show every loop based at $\displaystyle x_0$ is path homotopic to the path $\displaystyle e: I\to X$ defined by $\displaystyle e(x)=x_0$ for every $\displaystyle x\in X$

Let $\displaystyle \alpha$ be a loop based at fixed $\displaystyle x_0 \X$. $\displaystyle X$ is contractible implies there exists a homotopy $\displaystyle H: X \times I \to X$ such that $\displaystyle H(x,0)=x$ and $\displaystyle H(x,1)=x_0$. I tried to construct a path homotopy from $\displaystyle \alpha$ to $\displaystyle e$ using H, but I don't think mine works. I define $\displaystyle G(x,t)= H(\alpha(x),t)$. Then $\displaystyle G(x,0)=H(\alpha(x),0)=\alpha(x)$ and $\displaystyle G(x,1)=H(\alpha(x),1)=x_0$. But this is only a homotopy, I want $\displaystyle H(0,t)=H(1,t)=x_0$. I can't get it with this map.