Show that given a collection \(\mathcal{C}\) of spaces, the relation of homotopy equivalence is an equivalence relation on \(\mathcal{C}\).
Solution
Reflexivity: every space \(X\) is homotopy equivalent to itself via \(\operatorname{id}_X\).
Symmetry: if \(f:X\to Y\) and \(g:Y\to X\) satisfy \[ gf\simeq\operatorname{id}_X,\qquad fg\simeq\operatorname{id}_Y, \] then the same equations show that \(g\) is a homotopy equivalence from \(Y\) to \(X\) with homotopy inverse \(f\).
Transitivity: suppose \(X\simeq Y\) via \(f:X\to Y\), \(g:Y\to X\), and \(Y\simeq Z\) via \(u:Y\to Z\), \(v:Z\to Y\). Then \(u f:X\to Z\) and \(g v:Z\to X\) satisfy \[ (gv)(uf)=g(vu)f\simeq g\,\operatorname{id}_Y f=gf\simeq\operatorname{id}_X, \] and similarly \[ (uf)(gv)=u(fg)v\simeq u\,\operatorname{id}_Yv=uv\simeq\operatorname{id}_Z. \] Thus \(X\simeq Z\). Hence homotopy equivalence is an equivalence relation.