I am trying to understand the proof of Lemma 3.0.15 of [this paper](http://arxiv.org/abs/1312.7164v1) (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?
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> 


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).



comparison of truncations
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.


