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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

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Exercise 58.5

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Definition of fundamental group

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Definition of simply connected

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The context is the claim made here in Example 70.1

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

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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!

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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)...

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    $\begingroup$ OK. I saw I needed to show simple-connectedness. More thought needed. $\endgroup$ – Henno Brandsma Nov 11 '18 at 13:35
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    $\begingroup$ It comes for free as long as your contraction is a deformation retraction onto $x_0$ (these are the assumptions in your first paragraph). In favorable situations, we can modify a contraction to a deformation retraction onto a point, but this is not always guaranteed. (It is easy to prove, for instance, if you replace $X$ with $X \cup_{x_0} I$ - that is, $X$ with a whisker attached along $x_0$. Now you need to show these have the same pointed homotopy type, which requires some demands on the space.) $\endgroup$ – user98602 Nov 13 '18 at 16:34
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    $\begingroup$ On the other hand, here is an argument that implies that you do not need the contraction to preserve the basepoint. 1) Free homotopy classes of loops in $X$ are identified with conjugacy classes in $\pi_1(X, x_0)$. 2) A contractible space has one free homotopy class of loop. 3) A group with only one conjugacy class is trivial, because the identity element is not conjugate to anything else. $\endgroup$ – user98602 Nov 13 '18 at 16:35
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    $\begingroup$ @MikeMiller For a deformation retraction I would need $H(x_0,t)=x_0$ for all $t$, which was not in my first paragraph. If we would have it we'd be done straight away. But IIRC there are spaces $X$ such that $1_X$ is homotopic to the constant map $x_0$ without $x_0$ being a deformation retract of $X$. $\endgroup$ – Henno Brandsma Nov 13 '18 at 16:43
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    $\begingroup$ I made a mistake in reading your definition. My apologies. There absolutely are spaces that admit a contraction without admitting a deformation retraction to a point (easier to cook up examples so that you can't d.r. to a specific point, but there are some that you cannot d.r. to any point). $\endgroup$ – user98602 Nov 13 '18 at 16:56

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