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I'm studying P. Schapira's notes Algebra and Topology, available online here, and I'm having trouble understanding sheaves associated with locally closed subsets, in particular constant sheaves.

For example, the following is Exercise 5.5:

Let $X = \mathbb{R}^2$, $Y = \mathbb{R}$, $S = \{(x, y) \in X \mid x y \ge 1 \}$, and let $f : X \rightarrow Y$ be the map $(x, y) \mapsto y$. Calculate $f_* k_{XS}$.

Unravelling the definitions, I see that for $U \subseteq Y$ open, $$ (f_* k_{XS})(U) = k_{XS} (\mathbb{R}\times U) = (i_{S*} i_S^{-1} k_X) (\mathbb{R} \times U) = (i_S^{-1} k_X) ((\mathbb{R} \times U)\cap S)$$ but here I'm stuck. The problem is that I don't really know how to deal with the functor $i_S^{-1}$ (unless it appears in a context where I can use the adjunction $i_S^{-1} \dashv i_{S*}$).

I'd be really grateful if someone could help me solve the above exercise, and maybe give me some insight on how to handle this kind of problems in general. Thank you.

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You are doing fine - $(i^{-1}_Sk_X)((\mathbb{R} \times U \cap S)$ is fine. Now consider two cases: if U is an small interval containing $0$, or if it doesn't. – ykm Sep 23 '13 at 4:32
Ok, let me see if I understand. If $U$ is an interval and $0 \in U$ then $V = (\mathbb{R} \times U) \cap S$ has two connected components, $V_1$ and $V_2$, so I can write $(i_S^{-1} k_X)(V) = (i_S^{-1}k_X)(V_1) \oplus (i_S^{-1}k_X)(V_2)$; otherwise $V = V_1$ is connected. In both cases $(i_S^{-1} k_X)(V_i) = k$ because any locally constant function on a connected set is constant. Am I right? – Luca Bressan Sep 23 '13 at 8:35
yes, that's correct. – ykm Sep 24 '13 at 2:33

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