Homology as a Derived Functor

Question

I was going through some exercises in Weibel, three of which were

• prove derived functors are universal $\delta$-functors
• prove (co)homology is a universal $\delta$-functor
• prove that if $T_\ast$ is a $\delta$-functor, and $T_n$ is (co)effacable for $n > 0$, then $T_\ast$ is universal

I ended up using the third statement to prove the second, but it got me thinking: can I use the first one to prove the second? I.e., is $H_\ast$ a derived functor? If so, it's gotta be the left-derived functor of $H_0$ (cause $L_0 F \cong F$).

Is this actually true? I can prove it's true when all the objects in the complex are projective, but the general case eludes me.

EDIT: I have managed to make negative progress. If $A$ is any chain complex, let $\cdots \to P_2 \to P_1 \to P_0 \to A \to 0$ be a projective resolution in $Ch(\mathcal A)$. We want to compute the homology of $F(P)$. But each $P_n$, being a projective object in $Ch(\mathcal A)$, is a split exact complex of projectives, so $F(P_n) = H_0(P_n) = 0$. Presumably, the derived functors of $H_0$ are not all zero (especially not the zeroth one!)

What am I doing wrong here?

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The proof I've got so far:

Let $F = H_0$, because it'll make notation a little clearer at times. We'll show that this is true when $A$ is a complex of projective objects.

Let $A$ be such a complex. Consider the complex $P = \mathrm{cone}(A[1])$ (i.e., it is $A_n \oplus A_{n+1}$ in degree $n$). Since $P$ is split-exact and made of projective objects, it's projective in $Ch(\mathcal A)$. So we start building the projective resolution of $A$. The kernel of $P \to A$ is just $A_{n+1}$, so $P[1]$ (which is $A_{n+1} \oplus A_{n+2}$ in degree $n$) surjects onto it in the obvious way (we gotta truncate at degree 0 though). Repeating this, we get that $\cdots \to P_2 \to P_1 \to P_0 \to A \to 0$ is a projective resolution of $A$.

Next we need to apply $F = H_0$ to $P_n$. Normally it'd be $0$, since cones are split-exact. But we truncated at $0$, so we get $\mathrm{coker} (P_{n, 1} \to P_{n, 0})$, which is $A_n$. So $F(P)$ is just $A$, and thus $L_i F(A) = H_i(F(P)) = H_i(A)$.

The whole thing looks like $\require{AMScd}$ \begin{CD} @. \vdots @.\vdots @. \vdots @. \vdots @. \\ @. @VVV @VVV @VVV @VVV @. \\ \cdots @>>> A_4 \oplus A_5 @>>> A_3 \oplus A_4 @>>> A_2 \oplus A_3 @>>> A_2 @>>> 0 \\ @. @VVV @VVV @VVV @VVV @. \\ \cdots @>>> A_3 \oplus A_4 @>>> A_2 \oplus A_3 @>>> A_1 \oplus A_2 @>>> A_1 @>>> 0 \\ @. @VVV @VVV @VVV @VVV @. \\ \cdots @>>> A_2 \oplus A_3 @>>> A_1 \oplus A_2 @>>> A_0 \oplus A_1 @>>> A_0 @>>> 0 \\ \end{CD}

Answer 1

Your error is that a projective object in $Ch(\mathcal{A})$ does not have to be an exact chain complex. For instance, a chain complex $P$ which is $0$ in all nonzero degrees but $P_0$ is a projective will be projective. Indeed, given an epimorphism $f:Z\to P$ from some other chain complex $Z$, the map $Z_0\to P_0$ splits since $P_0$ is projective, and that splitting gives a chain map $P\to Z$ which splits $f$.

You can make your argument work for general chain complexes using spectral sequences. Start with a chain complex $A$ and take a projective resolution $\cdots \to P_2 \to P_1 \to P_0 \to A \to 0.$ Now consider $$\cdots \to P_2 \to P_1 \to P_0 \to 0$$ as a double complex in $\mathcal{A}$, with the differentials of the $P_n$ going vertically and the maps $P_{n+1}\to P_n$ going horizontally. There are two different spectral sequences which converge to the total homology of the double complex. The first spectral sequence has as its $E^1$ page the homology of the rows. But the rows are exact except at the end, where their homology is $A$, so the $E^1$ page is just $A$ in the $0$th column and $0$ in all the other columns. The $E^2$ page then is the homology of the columns, which is just the homology of $A$. Afterwards the spectral sequence degenerates, so we conclude that the total homology of our double complex is naturally isomorphic to the homology of $A$.

The second spectral sequence starts by taking homology of the columns. Since each $P_n$ is projective, it is split-exact except in the last term as discussed above. So the $E^1$ page will have $H_0(P_n)$ in the $0$th row and will be trivial in all other rows. The $E^2$ page is then the homology of the rows, which exactly computes $L_nH_0(A)$ in the bottom row. Again, the spectral sequence degenerates after that.

So, since both spectral sequences converge to the total homology of the double complex, we get a natural isomorphism $H_n(A)\cong L_nH_0(A)$. Note that to conclude that they are isomorphic as $\delta$-functors, a bit more work is needed (you need to know that this natural isomorphism respects the connecting homomorphisms). That can also be proved via the double complex above: a short exact sequence in $Ch(\mathcal{A})$ gives a short exact sequence of double complexes, and thus a long exact sequence relating their homologies. Identifying the homology of each double complex with $H_n(-)$ or $L_nH_0(-)$ via the spectral sequences, you can chase some definitions to see that this identification respects the connecting homomorphisms.

Answer 2

$ \require{AMScd} \newcommand{\hor}[1]{{}_= #1} \newcommand{\ver}[1]{#1^\|} \newcommand{\donor}[1]{#1_\square} \newcommand{\recep}[1]{{}^\square #1} $

I've been reading up on spectral sequences, in order to understand Eric Wofsey's answer. Don't have a good grasp on them yet, but they certainly seem powerful.

While doing that reading, I ran into the "salamander lemma", which led me to an elementary proof of this. I'm surprised the salamander lemma isn't more commonly talked about, it seems very useful.

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Given a chain complex $A$, let $\cdots \to P_2 \to P_1 \to P_0 \to A$ be a projective resolution. Then, as Eric said, each $P_n$ is exact except maybe at $P_{n, 0}$. So let $Q_n = H_0(P_n)$, and $X = H_0(A)$, and consider the following diagram: \begin{CD} @. \vdots @. \vdots @. \vdots @. \vdots \\ @. @VVV @VVV @VVV @VVV \\ \cdots @>>> P_{2,2} @>>> P_{1,2} @>>> P_{0,2} @>>> A_0 @>>> 0 \\ @. @VVV @VVV @VVV @VVV \\ \cdots @>>> P_{2,1} @>>> P_{1,1} @>>> P_{0,1} @>>> A_0 @>>> 0 \\ @. @VVV @VVV @VVV @VVV \\ \cdots @>>> P_{2,0} @>>> P_{1,0} @>>> P_{0,0} @>>> A_0 @>>> 0 \\ @. @VVV @VVV @VVV @VVV \\ \cdots @>>> Q_2 @>>> Q_1 @>>> Q_0 @>>> X @>>> 0 \\ @. @VVV @VVV @VVV @VVV \\ @. 0 @. 0 @. 0 @. 0 \\ \end{CD}

Every column except maybe the last, and every row except maybe the last, is exact.

The salamander on $Q_{n+1} \to Q_n$ gives the exact sequence $$ \donor{(P_{n+1,0})} \to \hor{(Q_{n+1})} \to \donor{(Q_{n+1})} \to \recep{(Q_n)} \to \hor{(Q_n)} \to \recep{0} $$

Since $P_{n+1,0} \to Q_{n+1}$ is surjective, $\donor{(Q_{n+1})} = 0$, and so $\recep{(Q_n)} \cong \hor{(Q_n)}$.

Since the relevant rows and columns are exact, the extramural maps form a chain of isomorphisms: $$ \recep{(Q_n)} \cong \donor{(P_{n,0})} \cong \recep{(P_{n-1,0})} \cong \donor{(P_{n-1,1})} \cong \cdots \cong \donor{(P_{0,n})} \cong \recep{(A_n)} $$

The salamander on $A_{n+1} \to A_n$ gives the exact sequence: $$ \donor{(P_{0,n+1})} \to \ver{(A_{n+1})} \to \donor{(A_{n+1})} \to \recep{(A_n)} \to \ver{(A_n)} \to \recep{0} $$

And because $P_{0,n+1} \to A_{n+1}$ is surjective, $\donor{(A_{n+1})} = 0$, and so $\recep{(A_n)} \cong \ver{(A_n)}$.

This chain of isomorphisms tells us that the horizontal homology at $Q_n$ is naturally isomorphic to the vertical homology at $A_n$. But the horizontal homology at $Q_n$ is exactly $L_n (H_0)$ of $A$.

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I don't want to think about how to show that this respects the connecting $\delta$ homomorphisms.