Let \(X\) be homotopy equivalent to a finite simplicial complex and let \(Y\) be homotopy equivalent to a finite or countably infinite simplicial complex. Using the simplicial approximation theorem, show that there are at most countably many homotopy classes of maps \(X \to Y\).
Solution
Choose homotopy equivalences \[ X\simeq K, \qquad Y\simeq L, \] where \(K\) is a finite simplicial complex and \(L\) is finite or countable.
1. For each \(r\ge0\), there are only countably many simplicial maps\[ \operatorname{sd}^rK\to L. \]
The complex \(\operatorname{sd}^rK\) has finitely many vertices, while \(L\) has at most countably many vertices. A simplicial map is determined by its map on vertices. There are only countably many maps from a finite set to a countable set, hence only countably many simplicial maps.
2. The union, over all \(r\ge0\), of the sets of such simplicial maps is countable.
It is a countable union of countable sets.
3. Every homotopy class of maps \(K\to L\) is represented by one of these simplicial maps.
By the simplicial approximation theorem, every continuous map \(K\to L\) is homotopic to a simplicial map after some iterated barycentric subdivision of the finite domain \(K\).
4. Therefore there are at most countably many homotopy classes of maps \(X\to Y\).
Composition with fixed homotopy equivalences and their homotopy inverses gives a bijection \[ [X,Y]\cong[K,L]. \] By <1>2–<1>3 the latter set is at most countable.