Let $\mathcal{A}$ be a "good" abelian category (The category I really care about is $R$-Mod for some noncommutative noetherian algebra $R$ over some field $k$, but I think the question still makes sense if $\mathcal{A}$ is an Ab-3 category with injective envelops), and $\mathcal{T}$ be a localizing subcategory of $\mathcal{A}$. Let $\tau\colon \mathcal{A}\to\mathcal{T}$ be the functor sending any object of $\mathcal{A}$ to its largest subobject in $\mathcal{T}$. An object $A\in\mathcal{A}$ is said to be torsion if $A\in\mathcal{T}$ (or $\tau(A)=A$), and is said to be torsion-free if $\tau(A)=0$.
Then the torsion theory tells us an object $N\in\mathcal{A}$ is torsion-free if $\text{Hom}_{\cal A}(M, N)=0$ for every torsion object $M\in{\cal T}$. Since otherwise $\tau(N)$ is a non-zero torsion subobject of $M$, and therefore $\text{Hom}_{\cal A}(M, N)$ will be non-zero.
My question is: Is there a similar proposition in derived category? Precisely, let $N\in D^b({\cal A})$ be a complex in the bounded derived category of $\cal A$, such that $$ \text{Hom}_{D({\cal A})}(M, N)=0 $$ for any complex $M\in D^b_{{\cal T}}({\cal A})$ in the bounded derived category of $\cal A$ with cohomologies in $\cal T$. Does $R\tau(N)=0$ always hold?
Here's what I've tried so far: If the opposite holds, i.e. $R\tau(N)\neq 0$, then there is a $n$ such that $R^n\tau(N)\neq 0$. Let $M=\sigma^{\leq n}(R\tau(N))$ be the canonical truncation of $R\tau(N)$. Then $M\in D^b_{{\cal T}}({\cal A})$ and I want to show that the canonical map $i\colon M=\sigma^{\leq n}(R\tau(N))\to R\tau(N)\to N$ is zero in $D({\cal A})$, where the map $R\tau(N)\to N$ is constructed as follows: Replace $N$ by its injective resolution $I$, and then it's clear that $\tau(I)$ is a subcomplex of $I$. I've guessed that $i$ must induce non-zero map on cohomologies, but it is not the case, and I can't go any further.
If $R^n\tau(N)\neq0$, let $X$ be the kernel of the differential $\tau I_n\to \tau I_{n+1}$ and $\alpha:X\to \tau I_n$ the natural inclusion.
Since $\text{Hom}_{D(\mathcal{A})}(X[-n],N)=0$, the composition $X\to \tau I_n\to I_n$ of $\alpha$ with the natural map $\tau I_n\to I_n$ factors through the differential $I_{n-1}\to I_n$.
The resulting map $X\to I_{n-1}$ factors uniquely through the natural map $\tau I_{n-1}\to I_{n-1}$, since $X\in\mathcal{T}$.
Hence the kernel of the differential $\tau I_n\to\tau I_{n+1}$ is contained in the image of $\tau I_{n-1}\to\tau I_n$, contradicting the fact that $R^n\tau(N)\neq0$.