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A manifold $M$ is simply connected if for every pair of 1-cubes $c_1,c_2: [0,1]\rightarrow M$ with
$c_1(0) = c_1(1) = c_2(0) = c_2(1) = t$
there is a 2-cube $b$ such that
1) $b(1,0) = c_1$ and $b(1,1) = c_2$
2) for all $p$ in $[0,1]$, $b(2,0)(p) = b(2,1)(p) = t$

My question is, if I'm given a manifold M,
for example choose $M = S^2 = \{(x,y,z)\in \mathbb{R}^3: x^2 + y^2 + z^2 = 1\}$
how do I prove that it is simply connected?

Also, how would you prove that a manifold is not simply connected?

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You apply the definition. Where are you having trouble? – Qiaochu Yuan Nov 16 '10 at 17:43
You need the path to not be an onto function $[0,1] \to S^2$ to do that. Space-filling curves are the bane of the topological category. :) – Ryan Budney Nov 16 '10 at 17:53
@JimJones: since you've tagged this differential geometry perhaps you're assuming your paths are smooth paths? The theorem which states that smooth paths can not be onto $n$-manifolds for $n \geq 2$ is called Sard's Theorem. – Ryan Budney Nov 16 '10 at 18:19
Saying $1$-cube is an amazingly effective way of breaking the connection with the intuitive content of the concept! – Mariano Suárez-Alvarez Nov 16 '10 at 19:14
@Mariano: Dear Mariano, Nicely put! I had the same reaction when I read this particular formulation of the definition. – Matt E Nov 17 '10 at 4:01

For the 2-sphere, it isn't hard to show it directly. However, in general, you'll need more powerful tools. A manifold $M$ is simply-connected if it is path-connected and if $\pi_1(M) = 1$. The standard tool to compute $\pi_1$ is the van Kampen theorem.

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Proving that a manifold $M$ is not simply connected (if true) is more difficult than proving that $M$ is simply connected (if true), because the first requires an "impossibility proof" (like proving that you cannot trisect an angle with ruler and compass). For an impossibility proof you need a "higher theory". E.g., in order to prove that an annulus $R$ in $\mathbb R^2$ is not simply connected you show that $R$ carries a closed vector field (namely $\nabla\arg $) which is not the gradient of a globally defined function. Such a thing would not exist on a simply connected domain.

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Not necessarily. Exhibiting a nontrivial connected covering space of a manifold will show immediately that it's not simply connected. – Robin Chapman Nov 17 '10 at 16:14

I believe that there is a nice way to study this problem via the language of simplicial complexes and triangulations.

Let us triangulate $S^1$ and $S^2$ by the simplicial complexes $K$ and $L$, respectively, where $K$ consists of the proper faces of a $2$-simplex (i.e., a triangle) and $L$ consists of the proper faces of a $3$-simplex (i.e., a tetrahedron). If $h:S^1\to S^2$ is a continuous map, then we can apply the finite simplicial approximation theorem to conclude that there is a simplicial map $f:K\to L$ such that $h$ is homotopic to $f$. However, $f$ maps $K$ into the $1$-skeleton of $L$, and this is certainly a proper subspace of $L$. In particular, $h:S^1\to S^2$ is homotopic to a continuous function $f:S^1\to S^2 - p$ where $p$ is a point of $S^2$. Since $S^2 - p$ is homeomorphic to the contractible space $R^2$ (via stereographic projection), it follows that $h$ is homotopic to a constant map. Therefore, $\pi_1(S^2)=0$.

In fact, the technique I have used in the previous paragraph is more widely applicable and can be used to prove the following fact, which I leave as an exercise:

Exercise 1 Prove that the $i$th homotopy group of the $n$-sphere, $\pi_i(S^n)$, is trivial for all $0\leq i < n$ and all positive integers $n$.

I hope this helps!

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