For an orientable n-manifold $M$ with boundary, we know that there is a Lefchetz theorem: $H^{k}(M)\cong(H^{n-k}(M,\partial M))^*$. Now if we consider a representation $\rho$ of the fundamental group of $M$ (for simplicity, we can assume that $\rho$ is trivial when restrict to the boundary). Is there a similar relationship between $H^{k}(M;\rho)$ and $H^{n-k}(M,\partial M;\rho)$?

In three dimensional topology, there is a famous "half die half alive" lemma. Which says when we map the first q-coefficient homology group of the boundary into the 3-manifold, the rank of the kernal is the genus of the boundary. This is a consequence of Lefchetz Duality. Is there a similar theorem when we consider twist coefficent? (Again, we can assume that the representation is trivial when restrict to the boundary).

For an orientable n-manifold $M$ with boundary, we know that there is a Lefschetz theorem: $H^{k}(M)\cong(H^{n-k}(M,\partial M))^*$. Now if we consider a representation $\rho$ of the fundamental group of $M$ (for simplicity, we can assume that $\rho$ is trivial when restricted to the boundary). Is there a similar relationship between $H^{k}(M;\rho)$ and $H^{n-k}(M,\partial M;\rho)$?

In three dimensional topology, there is a famous "half die half alive" lemma. Which says when we map the first q-coefficient homology group of the boundary into the 3-manifold, the rank of the kernel is the genus of the boundary. This is a consequence of Lefschetz Duality. Is there a similar theorem when we consider twisted coefficients? (Again, we can assume that the representation is trivial when restrict to the boundary).