(May 2016) Let \(X=S^1\times S^1\), also thought of as the standard quotient of the unit square \([0,1]\times[0,1]\), and let \(A=\{(x,x):x\in S^1\}\) be the diagonal of \(X\). Show that \(A\) is a retract of \(X\), but not a deformation retract of \(X\).
Solution
Write points of the torus as \((z,w)\in S^1\times S^1\). Define \[ r:S^1\times S^1\longrightarrow A, \qquad r(z,w)=(z,z). \] This map is continuous and, for every \((z,z)\in A\), \[ r(z,z)=(z,z). \] Hence \(r\) is a retraction of \(X\) onto \(A\).
Suppose \(A\) were a deformation retract of \(X\). Then the inclusion \[ i:A\hookrightarrow X \] would be a homotopy equivalence and therefore induce an isomorphism on fundamental groups. But \[ \pi_1(A)\cong\mathbb Z, \qquad \pi_1(X)\cong\mathbb Z^2, \] and, under these identifications, the diagonal inclusion induces \[ i_*:\mathbb Z\to\mathbb Z^2, \qquad n\longmapsto(n,n), \] which is not surjective. This contradiction shows that \(A\) is not a deformation retract of \(X\).