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Let $\mathcal{A}$ be an abelian category with enough projectives, $F:\mathcal{A} \to \mathcal{B}$ be a right exact functor. The left hyper-derived functors of $F$, $\mathbb{L}_i F$ are defined by: for every $\mathcal{A}$-chain complex $A$, let $P\to A$ be a Cartan-Eilenberg (projective) resolution, then $\mathbb{L}_i F(A)=H_i(Tot^{\oplus}(F(P)))$. $\mathbb{L}_i F$ is a functor from $\mathcal{Ch(A)}$, the category of chain complexes in $\mathcal{A}$, to $\mathcal{B}$.

Lemma 5.7.5 in Weibel ("Introduction to Homological Algebra") states that

If $0 \to A \to B \to C \to 0$ is a short exact sequence of bounded below complexes, then there is a long exact sequence $\cdots \to \mathbb{L}_{i+1}F(C) \to \mathbb{L}_iF(A) \to \mathbb{L}_iF(B) \to \mathbb{L}_iF(C) \to \cdots$

Weibel proves this by evoking the exercise above which is

Show that the functors $\mathbb{L}_i F$ restricted to $\mathcal{Ch_{\geq 0} (A)}$, the category of chain complexes with $A_p=0$ for $p < 0$, are the left derived functors of the right exact functor $H_0 F$.

However I am having difficulty seeing how the statement of the exercise even makes sense. It is proved in an earlier exercise that $\mathcal{Ch (A)}$ has enough projectives, but the proof breaks down for $\mathcal{Ch_{\geq 0} (A)}$. In fact if I am correct in my calculations, a spectral sequence argument shows that $\mathcal{Ch_{\geq 0} (A)}$ cannot have enough projectives. So we cannot talk about derived functors in the latter category.

If the derived functors are taken in the larger category $\mathcal{Ch (A)}$ (which if this is what the exercise meant, can be used to prove the lemma also), the question still doesn't seem to be true. This is because projective objects in $\mathcal{Ch (A)}$ are split exact chain complexes of projectives, so $H_0 F$ sends all these to zero, showing that the derived functors are all zero.

This exercise has been asked on this site before (The hyper-derived functors $\mathbb L_\bullet F$ are just derived functors of $H_0F$?), but the accepted answer there assumes the existence of enough projectives in $\mathcal{Ch_{\geq 0} (A)}$, which I don't think is true, as said above.

I have also tried to prove Lemma 5.7.5 directly, but given a short exact sequence of chain complexes $0 \to A \to B \to C \to 0$, I cannot construct Cartan-Eilenberg resolutions $P,Q,R$ of $A,B,C$ respectively, such that they fit into a short exact sequence $0 \to P \to Q \to R \to 0$. Namely I am having difficulty meeting the requirement that $H(Q,d^h)$ be a projective resolution of $H(B)$.

Any hints/references on how to prove the lemma/solve the exercise is greatly appreciated!

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  • $\begingroup$ As an aside, (AFAIK) the requirement that there are "enough projectives" are is a technical detail that greatly simplifies the theory, not an absolute requirement. $\endgroup$ – Hurkyl Jul 7 '17 at 2:00
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Here is an argument showing that $\mathcal{Ch}_{\geq 0} (\mathcal{A})$ has enough projectives if $\mathcal{A}$ has enough projectives. The inclusion functor $$ F : \mathcal{Ch}_{\geq 0} (\mathcal{A}) \to \mathcal{Ch} (\mathcal{A}) $$ has a left adjoint $$ G : \mathcal{Ch} (\mathcal{A}) \to \mathcal{Ch}_{\geq 0} (\mathcal{A}) $$ called the "stupid truncation" (see e.g. SP 0118). Since $G$ is a left adjoint to an exact functor, it preserves projective objects.

Let $A \in \mathcal{Ch}_{\geq 0} (\mathcal{A})$ be any object. Since $\mathcal{Ch}(\mathcal{A})$ has enough projectives, we can find a projective object $P \in \mathcal{Ch} (\mathcal{A})$ and a surjection $\varphi : P \to F(A)$. The morphism $G(P) \to A$ corresponding to $\varphi$ via the adjunction stated above is surjective, and $G(P)$ is projective by the above argument.


Explicitly, given a chain complex $A_{\bullet}$ where $A_{i} = 0$ if $i < 0$, we can find $P_{\bullet} \in \mathcal{Ch}(\mathcal{A})$ which is projective in $\mathcal{Ch}(\mathcal{A})$ and a surjection $\varphi_{\bullet} : P_{\bullet} \to A_{\bullet}$. Let $\sigma_{\geq 0}P_{\bullet}$ be the stupid truncation of $P_{\bullet}$, and let $\varphi'_{\bullet} : \sigma_{\geq 0}P_{\bullet} \to A_{\bullet}$ be the induced morphism of complexes. Note that $\varphi'_{\bullet}$ is still surjective. It remains to check that $\sigma_{\geq 0}P_{\bullet}$ is projective in $\mathcal{Ch}_{\geq 0} (\mathcal{A})$. Since $P_{\bullet}$ is a split exact chain complex of projectives, we have that $\sigma_{\geq 0}P_{\bullet}$ is the direct sum $\sigma_{\geq 0}P_{\bullet} \simeq P''_{\bullet} \oplus P'''_{\bullet}$ where $P''_{\bullet}$ is a split exact chain complex of projectives and $P'''_{\bullet}$ is a chain complex consisting of a single projective object of $\mathcal{A}$ in degree $0$. Both $P''_{\bullet}$ and $P'''_{\bullet}$ are projective in $\mathcal{Ch}_{\geq 0} (\mathcal{A})$.

(Note that if $P'''_{\bullet} \ne 0$ then $\sigma_{\geq 0}P_{\bullet}$ is not projective in $\mathcal{Ch} (\mathcal{A})$. This makes sense since the right adjoint of $F$ above, the "canonical truncation" functor $\tau_{\geq 0}A_{\bullet}$ of the above link, is not right exact.)

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  • $\begingroup$ Thank you very much in pointing out my mistake! Now I can just proceed as in the answer given in the linked question. $\endgroup$ – Tsang Jul 7 '17 at 2:31

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