Prove "Contractible implies simply connected" using tools in Munkres Topology. Context is theta-space. I've read this online, but I haven't seen this proved in Munkres Topology. Has it been? If so, where?
In any case, here is my attempt to show it using the tools given in the book. Please verify.
The book has 2 exercises on contractible spaces. One says contractible spaces are path connected and the other says a space is contractible if and only if the space has the same homotopy type of a one-point space. Since simply connected is defined as path connected and trivial fundamental group, I think that if a space has the same homotopy type as that of a one-point space, then the fundamental group of the space is trivial. Is that right? If so, I can work out the details myself.
Exercise 51.3b

Exercise 58.5

Definition of fundamental group

Definition of simply connected


The context is the claim made here in Example 70.1

 A: I thought of one. Use Corollary 58.6 with $h$ as the identity map. 

Since $h$ is nullhomotopic by definition of contractible, the induced homomorphism $h_{*}$ is both the identity isomorphism and the trivial homomorphism.
This means all loop classes are what they are mapped to, which is the identity class!
A: Straight from the definitions:
If $X$ is contractible this means there is a point $x_0 \in X$ and a continuous $H: X \times I \to X$ such that $H(x,0) = x$ for all $x \in X$, $H(x,1) = x_0$ for all $x \in X$. (i.e. The identity is homotopic to a constant map)
This means that if $x \in X$, the map $p_x: [0,1] \to X$ defined by $p_x(t) = H(X,t)$ is continuous (it's the composition of $H$ with the map $t \to (x,t)$ which are both continuous) and it's a path from $p_x(0) = H(x,0)= x$ to $p_x(1) = H(x,1) = x_0$. 
As we have a path from any $x$ to this fixed $x_0$, $X$ is path-connected: to get a path from $x$ to $y$, compose the path $p_x$ with the reverse path of $p_y$: 
$p(t) = H(x,2t)$ for $t \le \frac12$, $p(t) = H(y, 2-2t)$ for $t \ge \frac12$, which is continuous by the pasting lemma for $[0,\frac12]$ and $[\frac12,1]$, as the two definitions coincide for $t=\frac12$, as $H(x,1) = H(y,1) = x_0$. 
This defines a path from $x$ to $y$ directly from the homotopy $H$.
Now point (c) of exercise 3 implies that $[I,X]$ has but one element (the class of the constant map, which always is in any $[I,X]$) and this set already almost equal to $\pi_1(X,x_0)$ (the difference being that the homotopy must keep end-points fixed all the time). So we almost have that $X$ is simply connected: path-connected (see above) and $\pi_1(X,x_0)$ a single element for some $x_0$ (almost, keeping endpoints fixed is the only possible issue)...
