Derived equivalence in Residues and Duality

Question

Let $A'$ be a serre subcategry of $A$, let $A'$ has enough injectives and every injective object of $A'$ is also injective in $A$.Then the natural functor $c:D^+(A')\rightarrow D_{A'}^+(A)$ is eauvalence,where $D^+_{A'}(A)$ means full subcategory of $D^+(A)$ with ccohomology in $A'$.

$c$ is fully faithful is clear since every bounded below complex has injective resolution.This can be proved by Cartan-Eilenberg resolution.

My question is how to prove $c$ is dense? Since it is not equivalence of bounded derived category, it seems that it can not be proved by inducntion.

This equivalence is in Hartshorne's Residues and Duality (Chapter I Proposition 4.8).I can't understand his proof.

For the resolution of bounded below complex, his proof is as follows：

What is the meaning of $I^p/{im I^{p-1}}\oplus_{X^p}X^{p+1}$. Is this pushout? I can't find this notation in this book.It seems that proof is not so easy?

I found that I can't add pictures succsfully today.

Thank you in advance.

Answer 1

It's not an $\otimes$ its an $\oplus$. You have maps $f_p: X^p \to I^p, d_p:X^p\to X^{p+1}$ so the only reasonable definition is $$I^p \oplus_{X^p} X^{p+1} := (I^p \oplus X^{p+1}) / \text{im}\,(f_p\oplus d_p)$$ Just check the argument makes sense now - can you verify that $X^\bullet \to I^\bullet$ is a quasi-isomorphism? I think it's just a diagram chase - I didn't think too carefully, but you can take the rest as a sketch of a proof.

surjectivity

We take a cycle $y \in I^p$ and search for a cycle in $X^p$ mapping to it. To understand $y$, think about hte definition of the map $I^p \to I^{p+1}$ as a composition $$I^p \to I^p\oplus_{X^p}X^{p+1} \hookrightarrow I^{p+1}$$ so we ignore the inclusion and write $dy = (y',0)$. This is equivalent to 0 when $y'$ is in the image of $X^p \to I^p$, so we choose $x \mapsto y'$. The map $X^p \to I^p$ is defined by the composition $$X^p \overset{0\oplus d_p}\longrightarrow (I^{p-1}\oplus X^p)/ \text{im}(X^{p-1}) \hookrightarrow I^{p}$$ so $f_px = y$.

injectivity

Take a cycle $x\in X^p$ which maps to a boundary $y = f_px$ in $I^p$, we wish to show it is a boundary in $X^\bullet$. Again the image of $X^p$ in $I^p$ is contained in the direct sum/quotient, so we look just at the map

$$I^{p-1} \to I^{p-1}\oplus_{X^{p-1}}X^p$$

this means that the element $f_px = (0,x) \in I^{p-1}\oplus_{X^{p-1}}X^p$ is equivalent to an element of the form $(y',0)$ for $y' \in I^{p-1}$. What does $(0,x) \equiv (y',0)$ tell us? This means there is $x' \in X^{p-1}$ such that $dx' = x$ and $fx' = y$, the part we care about being that $dx' = x$, so yes $x$ is a boundary.

Answer 2

Suppose $X^i=0,\forall i＜0$. Consider:$ Z^0\rightarrow X^0\rightarrow B^1\rightarrow Z^1\rightarrow X^1\rightarrow B^2\rightarrow $.

Select injective resolution in $A'$ $Z^0\rightarrow I^0$.Then induced $f^0:X^0\rightarrow I^0$.Consider the induced morphism $B^1\rightarrow I^0/{Z^0}$. Take pushout $I^0/{Z^0}\oplus_{B^1}Z^1$.It lies in $A'$ since they pushout has same cokernel and $A'$ is Serre subcategory. Take injective resolution $I^0/{Z^0}\oplus_{B^1}Z^1\rightarrow I^1$. Then there is a induced morphism $X^1\rightarrow I^1$.

It is clear that the morphism $I^0\rightarrow I^1$ has kernel $Z^0$.

Next Consider the induced morphism $B^2\rightarrow I^2/(I^0/Z^0\oplus_{B^1}Z^1)$. Then take pushout.Step by step.

Every step we can find that $X$ and $I$ have same homology. And $Z^\cdot I,B^\cdot (I)\in A'$.