Let $M,N$ be $R$-modules, and let $P_{\bullet}\mapsto M$ be a projective resolution for $M$, and $N\mapsto I^{\bullet}$ an injective resolution for $N$.
My question is in regards to the isomorphism $H^{\bullet}(Hom(M,I^{\bullet})) \cong H^{\bullet}(Hom(P_{\bullet},N)).$ The way this is usullay done is by showing that $H^{\bullet}(Tot(Hom(P,I)))\cong H^{\bullet}(Hom(M,I^{\bullet})) \cong H^{\bullet}(Hom(P_{\bullet},N)).$
($Tot$ converts a double complex $C^{\bullet\bullet}$ into a chain complex which has in degree $n$, $\bigoplus_{p+q=n}C^{pq}$, and the differential map is obtained from the horizontal and vertical maps of the double complex, with minus signs for the vertical maps in odd columns).
So any proof of this usually involves considering a diagram of the form
$C^{\bullet\bullet}$ corresponds to the double complex in the first column, $A^{\bullet\bullet}$ corresponds to double complex $Hom(P_{\bullet},I^{\bullet})$, and $B^{\bullet\bullet}$ corresponds to the whole thing (so $A^{\bullet\bullet}$ with column $C^{\bullet\bullet}$ appended to the left.
Then the argument goes as follows:
Such a diagram gives rise to a Short exact sequence of double complexes of the form $$0 \mapsto A^{\bullet\bullet} \xrightarrow{f^{\bullet\bullet}} B^{\bullet\bullet} \xrightarrow{g^{\bullet\bullet}} C^{\bullet\bullet} \mapsto 0. $$
My issue is I am not exactly sure how to read this short exact sequence, as I am not really sure what the author means with this short exact sequence. For instance I am not quite sure what the R-modules corresponding to $A^{00}$, $B^{00}$ or $C^{00}$ are, and what the maps $f$ and $g$ are in this case.
If I had to take a guess, I would say that $C^{00} = Hom(M,I^{0})$, $B^{00}= Hom(M,I^{0})$, while $A^{00}=0$. Then the short exact sequence is $0\mapsto 0 \mapsto Hom(M,I^{0}) \xrightarrow{id} Hom(M,I^{0}) \mapsto 0$, with the second map being the identity.
Similarly, $C^{10}=0$, $B^{10}=Hom(P_{0},I^{0})$, and $A^{10}= Hom(P_{0},I^{0})$, which gives $$ 0 \mapsto Hom(P_{0},I^{0}) \xrightarrow{id} Hom(P_{0},I^{0}) \mapsto 0 \mapsto 0,$$ with the first map being the identity.
This doesn't seem right to me somehow. (same logic produces a short exact sequence $0\mapsto C^{\bullet\bullet} \mapsto B^{\bullet\bullet} \mapsto A^{\bullet\bullet}$, where the maps are either identity or zero, but this short exact sequence doesn't work in my case. )
Ultimately, I want to argue that $$0 \mapsto A^{\bullet\bullet} \xrightarrow{f^{\bullet\bullet}} B^{\bullet\bullet} \xrightarrow{g^{\bullet\bullet}} C^{\bullet\bullet} \mapsto 0, $$ induces a short exact sequence in $Tot$ since direct sums of SES is a SES again, and then argue the middle term is acyclic since its rows are exact, and the connecting homomorphism will in turn induce the correct isomorphism that I desire.
Any ideas how to read the SES? What the modules are in each degree and the corresponding maps?
Best Answer
You are reading it correctly. The thing that seems puzzling is that this is something non-trivial.
For that reason, let me adress the "same logic produces" remark. This is not true. $C^{\bullet \bullet}$ is not a subobject of $B^{\bullet \bullet}$. Implicitly, the maps of double complexes should respect (=be compatible with) both the horizontal and vertical differentials. If you try to consider $C^{\bullet \bullet}$ as a subobject of $B^{\bullet \bullet},$ the differentials in $C^{\bullet \bullet}$ to the right have to be zero, since they go into zero objects. But in $B^{\bullet \bullet},$ the differentials on the same spots are some nonzero maps. Therefore, the imagined inclusion $C^{\bullet \bullet} \rightarrow B^{\bullet \bullet}$ does not respect the differentials.
You can see for yourself that in the other direction there is no problem, that is, $A^{\bullet\bullet}$ is a sub-double complex of $B^{\bullet \bullet}$ and the quotient is isomorphic to $C^{\bullet \bullet}$.