I have copied the exercise below for reference. I was able to figure out how to do (a) and (d). Essentially the isomorphism in (d) is given by rank and determinant respectively. So let me focus on (b) and (c) in this post.
Please do not provide me with full solutions, but hints (and if solutions are easier, please hide the details for some of them).
For (b), the existence of the two step locally free resolution is doable. The statement that $\operatorname{det}(\psi(D))=\mathscr{L}(D)$ is also easy. What is stumping me is how to prove the independence of the choice of resolution. Checking that this defines a homomorphism follows from Hartshorne Exericse II.5.16(d) and by taking a simultaneous resolution.
There are two points in this long exercise that I am stuck.
Where I am stuck: How can I show that the definition is independent of choice of resolution?
An idea is to consider the two resolutions and use a result from homological algebra to extend the map $\mathscr{F}\rightarrow\mathscr{F}$ to a chain map $\mathscr{E_i}\rightarrow \mathscr{E_i'}$ between the two resolutions. This doesn't seem to do anything for me.
I also did some computations in the easy case where $X$ is affine and $\mathscr{F}$ is free, but then were unenlightening.
Where I am stuck #2:
How should I approach (c)?
I know there are answers circulating around, but I would like only a hint. Since we want a torsion sheaf at the end, I should try to to find $\mathscr{L}(D)^{\oplus r}$ such that the stalk at the generic point $\xi$ gives an isomorphism $\mathscr{L}(D)^{\oplus r}_\xi\rightarrow \mathscr{F}_\xi$. Since $\mathscr{F}_\xi$ has rank $r$, I really only need $\mathscr{L}(D)_\xi\cong \mathcal{O}_\xi$. So maybe part (b) with $\operatorname{det}(\psi(D))=\mathscr{L}(D)$ could be helpful here? Unfortunately, I do not know…
Exercise II.6.11 The Grothendieck Group of a Nonsingular Curve Let $X$ be a nonsingular curve over an algebraically closed field $k$. We will show that $K(X) \cong \operatorname{Pic} X \oplus \mathbb{Z}$, in several steps.
(a) For any divisor $D=\sum n_{i} P_{i}$ on $X$, let $\psi(D)=\sum n_{i} \gamma\left(k\left(P_{i}\right)\right) \in K(X)$, where $k\left(P_{i}\right)$ is the skyscraper sheaf $k$ at $P_{i}$ and 0 elsewhere. If $D$ is an effective divisor, let $\mathcal{O}_{D}$ be the structure sheaf of the associated subscheme of codimension 1 , and show that $\psi(D)=\gamma\left(\mathcal{O}_{D}\right)$. Then use (6.18) to show that for any $D, \psi(D)$ depends only on the linear equivalence class of $D$, so $\psi$ defines a homomorphism $\psi: \operatorname{Cl} X \rightarrow K(X)$.
(b) For any coherent sheaf $\mathscr{F}$ on $X$, show that there exist locally free sheaves $\mathscr{E}_{0}$ and $\mathscr{E}_{1}$ and an exact sequence $0 \rightarrow \mathscr{E}_{1} \rightarrow \mathscr{E}_{0} \rightarrow \mathscr{F} \rightarrow 0$. Let $r_{0}=$ rank $\mathscr{E}_{0}$, $r_{1}=\operatorname{rank} \mathscr{E}_{1}$, and define $\operatorname{det} \mathscr{F}=\left(\bigwedge^{r_{0}} \mathscr{E}_{0}\right) \otimes\left(\bigwedge^{r_{1}} \mathscr{E}_{1}\right)^{-1} \in \operatorname{Pic} X .$ Here $\bigwedge$ denotes the exterior power (Ex. 5.16). Show that det $\mathscr{F}$ is independent of the resolution chosen, and that it gives a homomorphism $\operatorname{det}: K(X) \rightarrow$ Pic $X$. Finally show that if $D$ is a divisor, then $\operatorname{det}(\psi(D))=\mathscr{L}(D)$.
(c) If $\mathscr{F}$ is any coherent sheaf of rank $r$, show that there is a divisor $D$ on $X$ and an exact sequence $0 \rightarrow \mathscr{L}(D)^{\oplus r} \rightarrow \mathscr{F} \rightarrow \mathscr{T} \rightarrow 0$, where $\mathscr{T}$ is a torsion sheaf. Conclude that if $\mathscr{F}$ is a sheaf of rank $r$, then $\gamma(\mathscr{F})-r \gamma\left(\mathcal{O}_{X}\right) \in \operatorname{Im} \psi$.
(d) Using the maps $\psi$, $\det$, $\operatorname{rank}$, and $1 \mapsto \gamma\left(\mathcal{O}_{X}\right)$ from $\mathbb{Z} \rightarrow K(X)$, show that $K(X) \cong$ $\operatorname{Pic} X \oplus \mathbb{Z}$.
Best Answer
You need to somehow connect your two resolutions $0\to \mathcal{E}_1\to \mathcal{E}_0\to \mathcal{F}\to 0$ and $0\to \mathcal{E}_1'\to \mathcal{E}_0'\to \mathcal{F}\to 0$. What if you considered $\mathcal{E}_0\oplus\mathcal{E}_0'\to \mathcal{F}$? Can you make any nice-looking homological algebra diagrams from this? You will need some homological algebra lemmas here, and they might not be the most commonly used ones.
You're on the right track by starting with an isomorphism $\mathcal{F}_\xi\cong \mathcal{O}_{X,\xi}^n$. You need to improve this twice: first, promote this to an isomorphism $\mathcal{O}_X^n|_U\to\mathcal{F}|_U$ for some open $U$; second, obtain an injection on the whole curve which after some fiddling will give you the first map in your exact sequence. Once you have the map in your exact sequence, write $\mathcal{T}$ in another way and play around with arithmetic in the Grothendieck group (you may find (a) helpful here).