Fix a class $a\in H_{p+q}(X,A)$. Then I want to show that the diagram $$ H^p(X,A) \rightarrow H^p(X)$$ $$ \downarrow a \cap\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow a \cap$$ $$ H_q(X) \rightarrow H_q (X,A)$$ is commutative(where $a \cap$ means cap product with $a$).
Here I am working with singular homology and singular cohomology.
I know that relative cap products $$ H_{p+q}(X,A) \times H^p(X,A) \rightarrow H_q(X),$$ $$ H_{p+q}(X,A) \times H^p(X) \rightarrow H_q(X,A)$$ are well defined. How to proceed with the proof?
Thank you for your help!