Problem: Let $X$ be a locally compact topological space, $i:Z \hookrightarrow X$ inclusion of a closed subspace, and $j:U \hookrightarrow X$ inclusion of the complement. I want to compute:
$$\text{RHom}(i_{!}k_{Z},j_{!}k_{U})$$
Moreover, I would like to be able to write a concise, rigorous proof of this result using basic machinery from the category of sheaves and derived categories. In particular, without having to deal with an explicit injective resolution of $k_U$. I know that this will, in general, depend on the topology of $Z$ and $X$. I would be satisfied with the answer in a simple case like $X = \mathbb{R}^n$, $Z = \{pt\}$.
Idea: By adjunction we have $\text{RHom}(i_{!}k_Z,j_!k_U) \cong \text{RHom}(k_Z,i^{!}j_{!}k_U)$. Now $j_{!}$ is extension by zero, and $i^{!}$ looks at sections relative to sections away from $Z$.
In particular, if we have an open set $V$ with $Z \subset V \subset \overline{V}$, and $\overline{V}$ deformation retracts onto $Z$, and $Z$ is contractible, we should have $\text{RHom}(k_Z,i^{!}j_{!}k_U) \cong H^{\bullet}(\overline{V};Z \cup \partial V)$- the relative cohomology.
So, is this true? If so, what would a rigorous proof look like?