Adar Kahiri

Basic Constructions

September 2020

Problem 1

Show that composition of paths satisfies the following cancellation property: If \(f_0 g_0 \simeq f_1 g_1\) and \(g_0 \simeq g_1\) then \(f_0 \simeq f_1\).

Proof. We first show that \(f_0\) and \(f_1\) have the same endpoints. Since \(f_0 g_0 \simeq f_1 g_1\), \(f_0(0) = f_1(0)\). Since \(g_0 \simeq g_1\), \(g_0(0) = g_1(0)\), and since we're dealing with path compositions, we know that \(f_0(1) = g_0(0)\) and \(f_1(1) = g_1(0)\), which means \(f_0(1) = f_1(1)\).

We must then show that there is a homotopy between \(f_0\) and \(f_1\). Our existing homotopy is \(h: I \times I \to X\), so then the desired homotopy is simply a reparametrization of the restriction \(h|[0, 1/2] \times I\) so that its domain is once again \(I \times I\).

This directly follows from the definition of path composition, whereby two composed paths are reparametrized so that they can each be 'traversed in half the time', so that they can both be traversed in unit time. As a result, for composable paths \(f\) and \(g\), \(f g (1/2) = f(1)\).


Problem 5 Show that for a space \(X\), the following three conditions are equivalent:

  1. Every map \(S^1 \to X\) is homotopic to a constant map, with image a point
  2. Every map \(S^1 \to X\) extends to a map \(D^2 \to X\)
  3. \(\pi_1 (X, x_0) = 0\) for all \(x_0 \in X\).

Proof. We show that \(1 \implies 2, 2 \implies 1, 1 \implies 3\), and \(3 \implies 1\).

\(1 \implies 2\). Given the map \(f: S^1 \to X\), our desired extended map is the composition of \(h_t: S^1 \to X\) with a map \(g: D^2 \to S^1 \times I\), where \(h_0 = f\) and \(h_1\) is the constant map. The map \(g: D^2 \to S^1 \times I\) can be defined like so (so that it is continuous and bijective):

\[g: ((1-r)\cos \theta, (1-r)\sin \theta) \to (\cos \theta, \sin \theta, r)\]

for \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2 \pi\) where \(r\) is the point's distance from the disk's center, and \(\theta\) is its angle from 0 on a circle with radius \(r\).

As might have noticed, this map can only be defined bijectively for \((S^1, 1)\) if \(f\) is nullhomotopic (homotopic to the constant map), for if it weren't nullhomotopic, then for any homotopy \(h_t\) of \(f\), the image of \(h_1\) would have more than one point, whereas the preimage of \(g\) would have only one point.

If it isn't intuitively clear why this is true, consider a homotopy \(h_t\) from a loop to a constant map. Now, instead of thinking of this homotopy as 'continuously deforming' a loop to a point, imagine all of the loops in the family \(h_t\) imbedded in \(X\) (which can a number of spaces, but for simplicity we'll say it is a sphere) at once. It should a little clearer how 'combining' all of the loops in \(h_t\) gives the image of a disk.

\(2 \implies 1\). Here we pretty much do the opposite of what we did above. We define a map \(g: S^1 \times I \to D^2\) like so:

\[g: (\cos \theta, \sin \theta, r) \to ((1-r)\cos \theta, (1-r)\sin \theta)\]

and we compose it with our extended map \(D^2 \to X\) to get our desired homotopy.

\(1 \implies 3\). If every map \(S^1 \to X\) is nullhomotopic, then every such map is homotopic to every other such map by transitivity, meaning there is only one homotopy class in \(X\). This is what it means for the fundamental group of a space to be zero.

\(3 \implies 1\). If the fundamental group of \(X\) is zero, then there iss only one homotopy class. The constant path must be in this homotopy class, for if it isn't, there would be another homotopy class containing it, leaving us with two homotopy classes total, which is a contradiction. Every loop is therefore homotopic to a constant path. \(\tag*{$\blacksquare$}\)