Why are Serre functors always exact?

Question

Let $k$ be a field and $\mathcal{T}$ be a $k$-linear triangulated category with finite dimensional spaces of morphisms. Bondal and Kapranov proved that every Serre functor on $\mathcal{T}$ is exact (Proposition 3.3 here). A different proof was given by Huybrechts in his Fourier-Mukai book (Proposition 1.46). The proof by Bondal and Kapranov is a bit technical and relies on the notion of a mattress. For now, I decided not to read it thoroughly. Concerning the one given by Huybrechts, I could not convince myself that it is complete. Could someone please shed a light on this, possibly providing a simple proof or explaining how the desired linear form in Huybrechts' proof is constructed?

Answer 1

(Notations as in Huybrechts' book.) Consider the maps $\mathrm{Hom}(C,S(B))\rightarrow\mathrm{Hom}(C,C_0)$ and $\mathrm{Hom}(A[1],C_0)\rightarrow\mathrm{Hom}(C,C_0)$, and let $I$ and $J$ be their images. Huybrechts tells you how to define linear forms $\Xi_I:I\rightarrow k$ ("condition $\mathrm{i}')$") and $\Xi_J:J\rightarrow k$ ("condition $\mathrm{ii}')$"), and shows that $\Xi_I|_{I\cap J}=\Xi_J|_{I\cap J}$. The latter condition says exactly that there is a linear form on $I+J$ restricting to $\Xi_I$ on $I$ and $\Xi_J$ on $J$; extending this form to all of $\mathrm{Hom}(C,C_0)$ gives the desired form $\Xi$ Serre dual to $\xi$.