Show that the wedge axiom for homology theories follows from the other axioms in the case of finite wedge sums.
Solution
It suffices to prove the binary case and then induct.
Let \(X\vee Y\) be the wedge at the common basepoint \(*\). Regard \(X\) as a subcomplex of \(X\vee Y\). Then \[ (X\vee Y)/X\cong Y. \]
1. The long exact sequence of the pair \((X\vee Y,X)\) splits into short exact sequences\[ 0\longrightarrow\widetilde h_n(X) \longrightarrow\widetilde h_n(X\vee Y) \longrightarrow\widetilde h_n(Y)\longrightarrow0. \]
By excision, \[ \widetilde h_n(X\vee Y,X)\cong\widetilde h_n(Y,* )\cong\widetilde h_n(Y). \] The retraction \(r:X\vee Y\to X\) collapsing \(Y\) to the wedge point satisfies \(r\circ i=\mathrm{id}_X\). Hence \(i_*:\widetilde h_n(X)\to\widetilde h_n(X\vee Y)\) is split injective. Exactness then forces the connecting map from the relative group to \(\widetilde h_{n-1}(X)\) to vanish, giving the displayed short exact sequence.
2. The sequence splits naturally, so\[ \widetilde h_n(X\vee Y)\cong \widetilde h_n(X)\oplus\widetilde h_n(Y). \]
The retraction \(X\vee Y\to Y\) collapsing \(X\) supplies a section of the quotient-induced map to the relative group. Equivalently, the two inclusions induce the direct-sum map, and the two retractions give its inverse.
Iterating the binary isomorphism gives the wedge axiom for every finite wedge.