I am reading a paper where the following trick is used:

To compute the left derived functors $L_{i}FM$ of a right-exact functor $F$ on an object $M$ in a certain abelian category, the authors construct a complex (not a resolution!) of acyclic objects, ending in $M$, say $A_{\bullet} \to M \to 0$, such that the homology of this complex is acyclic, and this homology gets killed by $F$. Thus, they claim, the left-derived functors can be computed from this complex.

Why does this claim follow? It seems like it should be easy enough, but I can't seem to wrap my head around it.


1 Answer 1


Compare with a projective resolution $P_\bullet\to M\to 0$. By projectivity, we obtain (from the identiy $M\to M$) a complex morphism $P_\bullet\to A_\bullet$, which induces $F(P_\bullet)\to F(A_\bullet)$. With a bit of diagram chasing you shold find that $H_\bullet(F(P_\bullet))$ is the same as $H_\bullet(F(A_\bullet))$.

A bit more explict: We can build a resolution of complexes $$\begin{matrix} &\downarrow && \downarrow&&\downarrow\\ 0\leftarrow &A_2&\leftarrow&P_{2,1}&\leftarrow&P_{2,2}&\leftarrow\\ &\downarrow && \downarrow&&\downarrow\\ 0\leftarrow &A_1&\leftarrow&P_{1,1}&\leftarrow&P_{1,2}&\leftarrow\\ &\downarrow && \downarrow&&\downarrow\\ 0\leftarrow &M&\leftarrow &P_{0,1}&\leftarrow&P_{0,2}&\leftarrow\\ &\downarrow && \downarrow&&\downarrow\\ &0&&0&&0 \end{matrix} $$ i.e. the $P_{i,j}$ are projective and all rows are exact. The downarrows are found recursively using projectivity so that all squares commute: If all down maps are called $f$ and all left maps $g$, then $f\circ g\colon P_{i,j}\to P_{i-1,j-1}$ maps to the image of $g\colon P_{i-1,j}\to P_{i-1,j-1}$ because $g\circ(f\circ g)=f\circ g\circ g=0$, hence $f\circ g$ factors through $P_{i-1,j}$, thus giving the next $f\colon P_{i,j}\to P_{i-1,j}$. We can apply $F$ and take direct sums across diagonals, i.e. let $B_k=\bigoplus_{i+j=k} FP_{i,j}$. Then $d:=(-1)^if+g$ makes this a complex. What interest us here, is that we can walk from the lower row to the left column by diagram chasing, thus finding that $H_\bullet(F(P_{0,\bullet}))=H_\bullet(F(A_\bullet))$. Indeed: Start with $x_0\in FP_{0,k}$ with $Fg(x_0)=0$. Then we find $y_1\in FP_{1,k}$ with $Ff(y_1)=x_0$. Since $Ff(Fg(y_1))=Fg(Ff(y_1))=0$, we find $y_2\in FP_{2,k-1}$ with $Ff(y_2)=y_1$, and so on until we end up with a cycle in $A_k$. Make yourself clear that the choices involved don't make a difference in the end (i.e. up to boundaries). Also, the chase can be performed just as well from the left column to the bottom row ...

  • $\begingroup$ Could you elaborate on the diagram chasing, this is exactly what my problem is. $\endgroup$ Dec 29, 2012 at 13:50
  • $\begingroup$ There's also a small typo: $H_{\bullet}(F(A_{\bullet}))$. $\endgroup$ Dec 30, 2012 at 10:38

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