# The Unapologetic Mathematician

## Simply-Connected Spaces

We say that a space is “simply-connected” if any closed curve $c$ with $c(0)=c(1)=p$ is homotopic to a constant curve that stays at the single point $p$. Intuitively, this means that any loop in the space can be “pulled tight” without getting caught up on any “holes”.

It turns out that this is equivalent to saying that every closed curve is the boundary of some parameterized square. Indeed, consider the following diagram I’ve drawn with the help of Geogebra:

This is a picture of the homotopy cylinder. The domain of a curve is the interval $[0,1]$, so the domain of the homotopy cylinder is the square $[0,1]$. I’ve labeled the sides to describe what the homotopy does to them: the lower edge $(x,0)$ follows the curve $c$; the upper edge $(x,1)$ is the constant point $p$; the two sides are also constant at $p$, meaning that we’re holding the curve’s ends fixed as we perform the homotopy. And so the homotopy is exactly a continuous (or smooth) map from the square into our space, and the boundary of the parameterized square is exactly the curve $c$. The converse — that any parameterized square can be homotoped to look like this — shouldn’t be hard to see.

So what does this mean for homology? Well, for cubic singular homology it means that $H_1(M)$ is exact if $M$ is simply-connected. Indeed, if $C$ is a closed $1$-chain, then it must be made up of a formal sum of curves. Any curve which isn’t already closed must have a start and an end, and the end must be the start of another curve, or else the boundary points of $C$ wouldn’t cancel off. We can break $C$ up — possibly non-uniquely — into a collection of closed curves, each of which is the boundary of some parameterized square, by the above argument. Thus $C$ is itself the boundary of this collection of squares; since all closed $1$-chains are exact, the first homology vanishes.

December 14, 2011 - Posted by | Differential Topology, Topology

1. […] Simply-Connected Spaces […]

2. […] here, and (b) the justification is that (for our purposes) the electric field is defined in some simply-connected region of space which has no “holes” one could wrap a path around. In fact, if the […]

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3. Hi! I dont see why the boundary of the square must be c. With the definition I see that the boundary must be c-p, seeing p as the constant singular cube.

Comment by Elvis | November 29, 2017 | Reply

• Try walking around the boundary of the square; first you’ll go along $c$, starting and ending at $p$. Then you’ll sit at $p$ for three times as long as it took to traverse $c$. That’s homotopy-equivalent to just traversing $c$.

Comment by John Armstrong | November 29, 2017 | Reply

• I see the intuition, but I mean the definition is the four 1-cubes (sides) of the square with a signal. Then we have 4 sides with different signals, and only two of them cancel each other. At the end we have a curve c and a constant map p. I dont see why my argument flaws, and I thought about it the entire night.

Comment by Elvis | November 29, 2017 | Reply

• These aren’t simplices. Parameterize the boundary of the square and map that to the space itself. That gives you a parameterized path in the space, which is homotopy-equivalent to the path that traces $c$ directly.

Comment by John Armstrong | November 29, 2017 | Reply