Bounded derived localizations embed fully faithfully

If cohomology vanishes above $b$, the map ${\tau }^{\le b}X\to X$ is a quasi-isomorphism. If it vanishes below $a$, $X\to {\tau }^{\ge a}X$ is one. When both bounds hold take $a\le b$ and combine them, obtaining a zigzag to ${\tau }^{\ge a}{\tau }^{\le b}X$, which is termwise bounded. Acyclic and zero complexes allow any such bounds.

For termwise bounded-above endpoints, any left-roof vertex is cohomologically bounded above, so replace it by its upper canonical truncation. This proves fullness from $D^{-}$. To test equality, first put two roofs at a common vertex and then use an equalizing denominator; upper-truncate this witness too. The equality already holds in $D^{-}$. For bounded-below endpoints use right roofs and lower truncation of the target vertices and equality witnesses.

For bounded endpoints first work in $D^{-}$, as just proved. Use right roofs there and lower-truncate their vertices and equality witnesses; they remain bounded above, so are now bounded. This proves full faithfulness of $D^b\to D^{-}\to D$. The cohomology functors show that every object in each essential image has the claimed bounds, while step 1.1 proves the reverse inclusion.

Shifts and cones preserve each cohomological boundedness condition by the long exact sequence. Cone triangles in the bounded models map to cone triangles in $D$, so each inclusion is exact and the described full subcategories are triangulated.