Sheaves with $\mathbb{R}$-constructible proper direct image closed under dualizing?

Question

I learned about $\mathbb{R}$-constructible sheaves very recently, so I hope this isn't a silly question:

Let $M$ be a real analytic manifold, and $U\subset M$ an open subanalytic subspace. Denote the embedding $U\hookrightarrow M$ with $j$. Let $D^b_{\mathbb{R}c}(k_M)$ for a field $k$ denote the objects of $D^b(k_M)$ with $\mathbb{R}$-constructible cohomologies. Finally, consider a $F\in D^b(k_U)$, such that $j_!(F)$ is in $D^b_{\mathbb{R}c}(k_M)$.

Now I wonder, if (or rather, why) it is true, that the dual $D_U(F)$ of $F$ has the same property, that $j_!(D_U(F))$ is in $D^b_{\mathbb{R}c}(k_M)$?

Edit: As $j$ is an open embedding, $j_!$ is of course exact, so I changed $Rj_!$ to $j_!$ in the above lines.

Any help would be highly appreciated!

Thanks in advance

Answer 1

I just came across my own question again and think meanwhile I can give an answer:

By definition (e.g. Definition 8.3.4 in sheaves on manifolds by M. Kashiwara and P. Schapira), a sheaf $G\in D^b(M)$ is $\mathbb{R}$-constructible if

1. there is a locally finite covering $M=\cup_\alpha M_\alpha$ of $M$ by subanalytic subsets such that $H^i(G)\vert_{M_\alpha}$ is locally constant for any $\alpha$ and $i$, and

2. for any $x\in M$, $G_x$ is a perfect complex.

In particular, for $G\in D^b_{\mathbb{R}c}(M)$ we have $G_U\in D^b_{\mathbb{R}c}(M)$ for the open subanalytic $U\subset M$ in question. Now set $G=j_!F\in D^b_{\mathbb{R}c}(M)$, then we have $D_MG\in D^b_{\mathbb{R}c}(M)$ and so $$j_!(D_UF)\simeq j_!(D_Uj^{-1}G)\simeq j_!(j^{-1}D_MG)\simeq (D_MG)_U\in D^b_{\mathbb{R}c}(M).$$