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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?2010-11-16
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    Technically your definition is incorrect, as your definition satisfies the property that the disjoint union of simply-connected spaces is simply-connected. The convention is that simply connected spaces need to be path connected *and* satisfy your property. In your example this isn't an issue, of course. To prove a sphere is simply connected you'll either need to use a theorem like Seifert-VanKampen, or you'll have to know something about approximating continuous functions by smooth or polynomial functions -- that any path can be homotoped to a path that isn't a space-filling curve.2010-11-16
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    @Ryan: in this case, can't you do it directly by stereographic projection?2010-11-16
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    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. :)2010-11-16
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    @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.2010-11-16
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    If I just choose a pair of one cubes that works, does that prove the space is simply connected? I'm having trouble because I don't know how to "choose" one cubes... what would they look like?2010-11-16
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    @JimJones: no. There's a reason why the definition says "for every pair of 1-cubes". So a constructive proof will require you submitting a "function" whose inputs are "pairs of 1-cubes" and the output is "the 2-cube satisfying the conditions (1) and (2)".2010-11-16
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    Saying *$1$-cube* is an amazingly effective way of breaking the connection with the intuitive content of the concept!2010-11-16
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    So I could create a function F: (set of one cubes)^2 -> (set of two cubes) by F(c1,c2)=b such that b1,0=c1, b1,1=c2 and b2,0(p)=b2,1(p)=p for all t in [0,1]?2010-11-16
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    What I really need is an example. I can't understand any of these concepts without examples.2010-11-16
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    If instead your question asked about the case of $\mathbb R^2$ instead of $S^2$, you could define your map on the 2-cube to be $tf(x)+(1-t)g(x)$ where $f$ and $g$ have are your maps from the interval which agree at the endpoints. So your 2-cube is being parametrized by $(t,x) \in [0,1]^2$.2010-11-16
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    So I could just say let c1(x) and c2(x) be two 1-cubes in S^2 with c1(0)=c1(1)=c2(0)=c2(1)=p, with the 2-cube being c1(t)cos(xpi/2)+c2(t)sin(xpi/2) for (t,x) in [0,1]^2?2010-11-17
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    @Mariano: Dear Mariano, Nicely put! I had the same reaction when I read this particular formulation of the definition.2010-11-17
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    @JimJones: what do you think?2010-11-17

3 Answers 3

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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.2010-11-17
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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!