comparison of truncations I am trying to understand the proof of Lemma 3.0.15 of this paper (Ben-Zvi, Nadler, Preygel - Integral transforms for coherent sheaves).
The context is of two triangulated categories $C,D$ with t-structures and a functor $F\colon C \to D$ which is left and right t-exact (up to shifts). 
Recall that $F$ is left exact up to a shift if $F(C^{\geq 0}) \subset D^{\geq l}$. Similarly for right exact (call $r$ the error term).
They claim that given an object $c \in C$ there exist natural maps 
$$ F(\tau_{>N} c ) \to \tau_{>N-r}F(c) $$
and
$$ \tau_{\leq N+l}F(c) \to F(\tau_{\leq N}c) $$
whose cones have good properties.

How are these maps defined?

They moreover assume C and D to be enhanced (ie they work with stable $\infty$-categories), complete and cocomplete; and they assume F to be conservative and (homotopy) limit and colimit preserving (although this might actually not be important for my question on its own).
 A: Are you sure about your definition of right and left t-exact up to a shift? I think the names are swapped. Let me answer assuming left t-exact up to a shift means $F(C_{\le 0}) \subset D_{\le l}$. (Wait! You write the t-structure with superscripts instead of subscripts: is that related to me feeling the definitions are swapped? Maybe by $C^{\le n}$ you mean what I would call $C_{\ge -n}$? --in which case we are in agreement.)
Here's how the maps are defined: apply $\tau_{\le N+l} \circ F$ to the canonical morphism $c \to \tau_{\le N} c$ to get a morphism $\tau_{\le N+l} F(c) \to \tau_{\le N+l}F(\tau_{\le N} c)$. Since $F$ satisfies $F(C_{\le 0}) \subset D_{\le l}$ and commutes with shifts (you said it is assumed to preserve finite limits and colimits), $F(C_{\le N}) \subset D_{\le N+l}$. So we have $F(\tau_{\le N} c)\in D_{\le N+l}$, and thus $\tau_{\le N+l}F(\tau_{\le N} c) \cong F(\tau_{\le N} c)$. The other map is defined similarly.
I took a look at the paper to see what is claimed about the fiber of that map, and I am not sure I get it. They say the fiber is $F(\tau_{>N}c)$ and this lies in $D_{\ge N-r}$, so that the map becomes an isomorphism when you apply $\tau_{\le N-r}$. It seems to me that the fiber is not $F(\tau_{>N}c)$, but that the conclusion about $\tau_{\le N-r}$ is true.
Let's try to show that. Start from the exact triangle $\tau_{>N}c \to c \to \tau_{\le N}c$. Applying $F$ results in an exact triangle (again because we assume $F$ commutes with finite limits). So the fiber of $F(c) \to F(\tau_{\le N} c)$ is $F(\tau_{>N}c)$, but then this means it won't also be the fiber of the map $\tau_{\le N+l} F(c) \to F(\tau_{\ge N}c)$ unless $F(c) \in D_{\le N+l}$ (when the maps are basically the same).
On the other hand, the claim about $\tau_{\le N-r}$ seems correct: $\tau_{\le N-r}$ will send an exact triangle in $D$ to a cofibre sequence in $D_{\le N-r}$ (because it is left adjoint to the inclusion $D_{\le N-r} \hookrightarrow D$). In the resulting cofibre sequence, $\tau_{\le N-r} F(\tau_{>N}c) \to \tau_{\le N-r} F(c) \to \tau_{\le N-r} F(\tau_{\le n} c)$, the first object is 0 because $F$ is right t-exact up to a shift of $-r$, which shows that the second map is an equivalence.
