Show that a CW complex is contractible if it is the union of an increasing sequence of subcomplexes \(X_1 \subset X_2 \subset \cdots\) such that each inclusion \(X_i \hookrightarrow X_{i+1}\) is nullhomotopic, a condition sometimes expressed by saying \(X_i\) is contractible in \(X_{i+1}\). An example is \(S^\infty\), or more generally the infinite suspension \(S^\infty X\) of any CW complex \(X\), the union of the iterated suspensions \(S^n X\).
Solution
First, \(X\) is path-connected. Given \(x,y\in X\), both lie in some \(X_i\); the nullhomotopy of \(X_i\hookrightarrow X_{i+1}\) supplies paths from their images to the same point of \(X_{i+1}\).
Now let \[ f:S^n\to X \] be any based map, \(n\ge1\). Since \(S^n\) is compact and \(X\) is a CW complex, \(f(S^n)\) is contained in a finite subcomplex of \(X\). Because the \(X_i\)’s are increasing and have union \(X\), this finite subcomplex is contained in some \(X_i\). Thus \(f\) factors through \(X_i\).
But the inclusion \[ X_i\hookrightarrow X_{i+1} \] is nullhomotopic, so the composite \(S^n\to X_i\to X_{i+1}\), and hence \(f:S^n\to X\), is nullhomotopic. Consequently \[ \pi_n(X)=0\qquad(n\ge1). \] The map \(X\to *\) therefore induces isomorphisms on all homotopy groups (and on \(\pi_0\)). Since both spaces have CW type, Whitehead’s theorem gives \[ \boxed{X\simeq *}. \] Thus \(X\) is contractible. In particular this applies to \(S^\infty\) and to infinite suspensions \(S^\infty X\).