In the book by Stöcker & Zieschang, the Poincaré duality is obtained by an isomorphism $\gamma \colon \operatorname{Hom}(C_q(K), \mathbb{Z}) \to C_{n-q}^{\ast}(K^{\prime})$, $\gamma(\varphi) = \widetilde{\chi}_{\cdot}(\varphi)\frown z^{\prime}$. $K$ is a simplicial complex, $K^{\prime}$ its first barycentric subdivision, $\chi_{\cdot}\colon C_q(K^{\prime}) \to C_q(K)$ is induced by the map that assigns a fixed vertex of $\sigma$ to a barycenter $\hat{\sigma}$ of $K^{\prime}$ and $\widetilde{\chi}$ is its dual under $\operatorname{Hom}(\cdot, \mathbb{Z})$. $z^{\prime}$ is a fundamental cycle of $K^{\prime}$.
From this isormorphism, the authors construct an isomorphism $\operatorname{Hom}(C_q(K,L), \mathbb{Z}) \to C_{n-q}^{\ast}(L_c^{\prime})$ by restriction, where $L$ is a subcomplex of $K$ and $L_c^{\prime}$ the simplicial complement of $L$ in $K^{\prime}$.
These two isomorphisms give rise to isomorphisms $H^q(K) \to H_{n-q}(C^{\ast}(K^{\prime}))$ and $H^q(K,L) \to H_{n-q}(C^{\ast}(L_c^{\prime}))$ which in the end lead to isomorphisms $H^q(M) \to H_{n-q}(M)$ and $H^q(M, A) \to H_{n-q}(M-A)$ for the underlying spaces $M=|K|$ and $A=|L|$ of the complexes (if $M$ is a manifold).
Now, the other way round, there's an isomorphism $H^q(M-A) \to H_{n-q}(M,A)$ which is constructed in a similar way by an isomorphism $\operatorname{Hom}(C_q^{\ast}(L_c^{\prime}), \mathbb{Z}) \to C_{n-q}(K,L)$. But this one is not given by a cap product as the one above, or is it? If it's not given in a exlicit way like with the cap product, how can I check that this map that maps dual basis elements $\widetilde{\sigma^{\ast}}$ to simplices $\sigma$ ($\sigma^{\ast}$ is the fundamental cycle of the oriented dual cell of $\sigma)$ maps cocycles to cycles and coboundaries to boundaries?