Does $X$ being path-connected and contractible mean the fundamental group of $X$ is trivial?

Question

I just started learning algebraic topology recently.I'm not sure if I got the idea correctly.Hope someone would check it.

My solution:Since $X$ is contractible and path-connected,then $X$ is simply connected.Thus $\pi_{1}(X)$ is trivial.

Does this work?

Answer 1

The fact that $X$ is simply connected is what you should prove. Note that $X$ is always path connected if it is contractible: suppose that $id_X \simeq c_{x}$ for some $x \in X$ via a homotopy

$$ H : X \times I \to X, $$

which satisfies $H_0 = id$ and $H_1 = c_{x}$. For any $y \in X$, the map $H(y,-)$ is a path from $y$ to $x$.

To show that $\pi_1(X)$ you can see that moreover we have $X \simeq *$,

Hint: there is a unique map $X \to *$. As for maps $* \to X$, they correspond to a selection of some point of $X$. There aren't many to choose. Using the fact that $X$ is contractible you can prove that both compositions are homotopic to the identities.