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Let $X$ be a contractible space (i.e., the identity map is homotopic to the constant map). Show that $X$ is simply connected.

Let $F$ be the homotopy between $\mathrm{id}_X$ and $x_0$, that is $F:X\times [0,1]\to X$ is a continuous map such that $$ F(x,0)=x,\quad F(x,1)=x_0$$ for all $x\in X$. Next let $f:[0,1]\to X$ be a loop based $x_0$, I need to show that $f$ is path homotopic to the constant path $e_{x_0}$, The continuous map $G:[0,1]\times [0,1]\to X$ defined by $G(s,t)=F(f(s),t))$ is a homotopy between $f$ and $e_{x_0}$, but it may not be a path homotopy (i.e., $G(0,t)=G(1,t)=x_0$ for all $t\in [0,1]$), how to get around this?

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  • $\begingroup$ Try this: If Y is contractible, then any two maps $X\to Y$ are homotopic (indeed they are null homotopic). $\endgroup$ – hamid kamali Mar 21 '16 at 13:01
  • $\begingroup$ Definitely the right idea, but in $\pi_1$ we use based homotopies, so there is some fiddling with basepoints required. Basically it's a special case of the fact that a homotopy equivalance induces an isomorphism on $\pi_1$. Try using basepoint change isomorphisms to "fix" the problem. $\endgroup$ – Justin Young Mar 21 '16 at 16:38
  • $\begingroup$ @JustinYoung I still want to get a path homotopy from the homopty $G$, I think it can be done. $\endgroup$ – Xiang Yu Mar 21 '16 at 16:43
  • $\begingroup$ Yeah, that's just unraveling the proof of the result I cited. Use conjugation by basepoint change path. $\endgroup$ – Justin Young Mar 21 '16 at 18:20
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Since $X$ is contractible there exists a homotopy $h=h_t: id_X \to x_0$ where $x_0 \in X$. Therefore, if we take any loop $\alpha: I \to X$ then $h|_{\alpha}$ takes $\alpha$ to $x_0$. Can you show $X$ is path-connected? You know $h_t$ takes every point in $X$ to $x_0$ therefore given any $x,y \in X$ then $h_{2t}(x) * h^{-1}_{2t-1}(y)$ is a path from $x$ to $y$ where $h^{-1}$ represents to reverse of the homotopy i.e $h^{-1}_0(y) = x_0$ and $h^{-1}_1(y) = y$.

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We can multiply $G$ by a homopoty $H$ to get around the obstacle. More precisely, let $H:[0,1]\times [0,1]\to X$ be the homopoty defined by $H(s,t)=G(0,st)$. We note that $H(1,t)=G(0,t)=G(1,t)=H(1,t) $ for all $t\in[0,1]$, thus for every $t\in[0,1]$, the product $H_t*G_t*\overline{H_t}$ can be defined, where $H_t:[0,1]\to X, s\mapsto H(s,t)$ is the path of $H$ at time $t$. $\overline{H_t}$ is the reverse path of $H_t$ (i.e., $\overline{H_t}(s)=H_t(1-s)$), the product of paths induces a product of homotopies. We can thus define a homotopy $K:[0,1]\times [0,1]\to X$ by $$K(s,t)=\begin{cases} H_t(4s)= H(4s,t)=G(0,4st),& s\in[0,1/4]\\ G_t(4s-1)=G(4s-1,t),&s\in[1/4,1/2]\\ \overline{H_t}(2s-1,t)=H(2-2s,t)=G(0,(2-2s)t),&s\in[1/2,1] \end{cases}$$ Note that $K$ is continuous by the pasting lemma.

Now the path $K(s,0)=H_{0}*G_{0}*\overline{H_0}=e_{x_0}*f*\overline{e_{x_0}}$ is path homotopic to $f$ (note that $[e_{x_0}*f*\overline{e_{x_0}}]=[e_{x_0}]*[f]*[\overline{e_{x_0}}]=[f]$). Similarily, the path $K(s,1)=H_{1}*G_{1}*\overline{H_1}=H_t*e_{x_0}*\overline{H_t}$ is path homotopic to $e_{x_0}$. We also have $K(0,t)=K(1,t)=G(0,0)=x_0$ for all $t\in[0,1]$, thus $K$ is a path homotopy between $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$, thus $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$ are path homotopic, by previous discussion, we see that $f$ and $e_{x_0}$ are also path homotopic.

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