union and sum induce isomorphism in homology I am reading Hatcher's book. I have some difficulty understanding the relative cup product. In the proof it says if A and B are open in X, the inclusion $C^n(X, A\cup B;R)\rightarrow C^n(X, A+B;R)$   induce isomorphisms on cohomology. I wonder why $C_n(A+B)\rightarrow C_n(A\cup B)  $induce an isomorphism on homology. Knowing this, I think by applying the universal coefficient theorem and the five lemma we can show the relative cohomology is isomorphic. Thanks for any hint!
 A: Since $C_n(A+B)$ is the group of chains of simplices in $A$ or in $B$, we could also write it as $C_n(A)+C_n(B)$. There is an obvious inclusion $i:C_n(A+B)\hookrightarrow C_n(A\cup B)$, and this map has a retraction $\rho:C_n(A\cup B)\to C_n(A+B)$ sending a simplex $\sigma$ in $A\cup B$ to the alternating sum of the simplices in some subdivision of $\sigma$ such that each simplex is in $A$ or in $B$. For the construction of this map, see the proof of proposition $2.21$ in Hatcher's book, where it's shown that $i\rho$ is chain homotopic to the identity. This implies that $i$ and $\rho$ are inverse to each other on the level of homology, so $i_*:H_*(A+B)\to H_*(A\cup B)$ is an isomorphism.
It thus also induces an isomorphism between cohomology groups $H^n(A\cup B;M)\to H^n(A+B;M)$, by the universal coefficient theorem. The canonical map between the short exact sequences $$0\to C_*(A+B)\to C_*(X)\to C_*(X,A+B)\to 0\quad  \text{and}\\ 0\to C_*(A\cup B)\to C_*(X)\to C_*(X,A\cup B)\to 0$$
induces a map between the long exact sequences 
$$\dots\to H^n(A\cup B;M)\to H^{n+1}(X,A\cup B;M)\to H^{n+1}(X;M)\to\dots$$
and 
$$\dots\to H^n(A+ B;M)\to H^{n+1}(X,A+ B;M)\to H^{n+1}(X;M)\to\dots$$
Applying the five lemma, we see that all maps are isomorphisms.
