Quasi-circle has trivial homotopy but is not contractible

Let \(Y\) be the quasi-circle of §1.3, Exercise 7. That exercise proves two facts we use:

1. every path in \(Y\) is contained in a finite truncation of the oscillating graph together with the vertical segment and the connecting arc, hence in a finite tree;
2. the quotient map \[ q:Y\to S^1 \] collapsing the vertical segment has no lift to \(\mathbb R\to S^1\).

We show first that every homotopy group of \(Y\) is trivial. The space \(Y\) is path-connected, so \(\pi_0(Y)\) is trivial as a pointed set. For \(n\ge1\), let \[ f:S^n\to Y. \] The sphere \(S^n\) is a compact, connected, locally connected metric space, hence a Peano continuum. Therefore there is a continuous surjection \[ r:I\twoheadrightarrow S^n. \] The composite \(f r:I\to Y\) is a path and has image exactly \(f(S^n)\). By the path-confinement result above, this image lies in a finite tree \(T\subset Y\). Hence \(f\) factors through the contractible space \(T\), so \(f\) is nullhomotopic. Thus \[ \boxed{\pi_n(Y)=0\quad\text{for all }n\ge1.} \]

Nevertheless \(Y\) is not contractible. If it were, the quotient map \(q:Y\to S^1\) would be nullhomotopic. A nullhomotopy of \(q\), together with a lift of the terminal constant map to \(\mathbb R\), lifts through the covering \(\mathbb R\to S^1\) by the homotopy lifting property. Restricting the lifted homotopy to the initial time would give a lift of \(q\), contradicting §1.3, Exercise 7.

Therefore \(Y\) is weakly contractible but not contractible. If \(Y\) had the homotopy type of a CW complex \(K\), then \(K\) would also have all homotopy groups trivial, so Whitehead’s theorem would make \(K\) contractible, hence \(Y\) contractible. Contradiction. Thus \(Y\) does not have CW homotopy type.