Show that the coproduct in the Hopf algebra \(H_*(X; R)\) dual to \(H^*(X; R)\) is induced by the diagonal map \(X \to X \times X\), \(x \mapsto (x, x)\).
Solution
Assume the degreewise finiteness hypotheses needed to identify homology and cohomology as graded duals. Let \[ D:X\to X\times X, \qquad D(x)=(x,x), \] be the diagonal map. Under the Künneth identification, write \[ D_*:H_*(X;R)\to H_*(X;R)\otimes H_*(X;R). \]
For \(z\in H_*(X;R)\) and cohomology classes \(\alpha,\beta\), naturality of the Kronecker pairing gives \[ \langle D_*z,\alpha\times\beta\rangle =\langle z,D^*(\alpha\times\beta)\rangle. \] But by the definition of cup product, \[ D^*(\alpha\times\beta)=\alpha\smile\beta. \] Hence \[ \langle D_*z,\alpha\times\beta\rangle =\langle z,\alpha\smile\beta\rangle. \] This identity says exactly that \(D_*\) is the linear dual of the cup-product multiplication \[ H^*(X;R)\otimes H^*(X;R)\to H^*(X;R). \] Therefore the coproduct in the Hopf algebra \(H_*(X;R)\) dual to \(H^*(X;R)\) is precisely the map induced by the diagonal.