2
$\begingroup$

Consider a cohomological delta functor $T^*:\mathscr{A}\to\mathscr{B}$ between abelian categories such that $T^i$ is an effaceable functor for all $i>0$, i.e. for all $i>0$, and for any $A\in\mathscr{A}$, there exists an injection $u_A:A\hookrightarrow M_A$ in $\mathscr{A}$ such that $T^i(u_A) = 0$. Grothendieck famously proved in his Tōhoku paper that this implies that $T^*$ is a universal $\delta^*$-functor, which he later on used to prove his Vanishing Theorem.

However, in his Tōhoku paper, Grothendieck skips over most of the fine details of the proof, and it's those details that I'm trying to fill in for myself. Here is my work so far:

Fix some $\delta^*$-functor $S:\mathscr{A}\to\mathscr{B}$, and suppose we are given $f^0:T^0\to S^0$. My general idea is to go by induction (as I see that Grothendieck does), and assume that for some $n\ge 1$, we have defined $f^i:T^i\to S^i$ for all $0\le i<n$ commuting with all connecting morphisms. Then it is sufficient to show that we can uniquely define $f^n:T^n\to S^n$ commuting with the connecting morphisms, i.e. such that for any exact sequence $$0\to A\to B\to C\to 0$$ in $\mathscr{A}$, the following square commutes: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T^{n-1}(C) & \ra{\delta^n} & T^n(A)\\ \da{f^{n-1}_C} & & \da{f^n_A}\\ S^{n-1}(C) & \ras{\delta^n} & S^n(A) \\ \end{array} $$

Now, my first thoughts are to fix the maps $u_A$, and use them to define a short exact sequence $$0\to A\xrightarrow{u_A}M_A\xrightarrow{p_A}P_A\to 0$$for all $A\in\mathscr{A}$. If we let this exact sequence induce the connecting maps ${^T\!\!\delta_A^n}:T^{n-1}(C)\to T^n(A)$ and ${^S\!\!\delta_A^n}:S^{n-1}(C)\to S^n(A)$, then basic diagram chasing gives us that there exists a unique $f^n_A:T^n(A)\to S^n(A)$ such that the following diagram commutes:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T^{n-1}(M_A) & \ra{T^{n-1}(p_A)} & T^{n-1}(P_A) & \ra{{^T\!\!\delta_A^n}} & T^n(A) & \ra{0} & T^n(M_A)\\ \da{f^{n-1}_{M_A}} & & \da{f^{n-1}_{P_A}} & & \da{f^n_A}\\ S^{n-1}(M_A) & \ra{S^{n-1}(p_A)} & S^{n-1}(P_A) & \ras{{^S\!\!\delta_A^n}} & S^n(A) \\ \end{array} $$

which of course gives us uniqueness. But how do we show naturality, and how do we show that the natural transformation $f^n:T^n\to S^n$ (once we've proved naturality) commutes with the connecting morphisms?

$\endgroup$
  • 1
    $\begingroup$ Such details can be found in these lecture notes. $\endgroup$ – Fred Rohrer Jan 25 '17 at 6:10
  • $\begingroup$ I can't find anything about universal functors in those lecture notes. $\endgroup$ – D. Wynter Jan 25 '17 at 16:43
  • $\begingroup$ Oh yes, you are right. (Still, the relevant argument is essentially contained in 8.21.) $\endgroup$ – Fred Rohrer Jan 25 '17 at 19:21
2
$\begingroup$

The key obervation is that if you have a map $\phi\colon A\to B$, then you can choose $M_A$ and $M_B$ so that this extends to $\tilde{\phi}\colon M_A\to M_B$: choose any $M_A$ you like, and then choose $M_B=M_{M_A\oplus B}/A$ where we embed $A$ in $M_A\oplus B$ by $(a,-\phi(a))$. The map $\tilde{\phi}\colon M_A\to M_B$ and inclusion of $B$ are induced by the obvious maps to $M_A\oplus B$. This allows us to transfer naturality from the lower degree.

For the connecting maps, I believe you can show that $M_B/M_A$ can be chosen as $M_C$ (look at the long exact sequence). By the 9-lemma, we also have a SES given by $0\to M_A/A\to M_B/B \to M_C/C\to 0$; commutation with the boundary map a degree down for this sequence gives it for $0\to A\to B \to C\to 0$. I don't blame Grothendieck for not writing this out; it hurts my head to try to picture the diagrams.

$\endgroup$
  • $\begingroup$ But how do we know that the map $f_A^n$ is independent of the choice of $M_A$? $\endgroup$ – D. Wynter Jan 25 '17 at 3:24
  • 1
    $\begingroup$ If $M_A$ and $\tilde{M}_A$ are both possibilities, you show that the natural map $A\to M_A\oplus \tilde{M}_A/\{a,-a)\mid a\in A\}$ also effaces, and then use the maps induced by the inclusions of $M_A$ and $\tilde{M}_A$ into this bigger object. $\endgroup$ – Ben Webster Jan 26 '17 at 1:35
  • 1
    $\begingroup$ Commutativity with connecting maps can be shown in a simpler way: for $0\to A\stackrel f\to B\to C\to0$ an exact sequence and $u_B:B\hookrightarrow M_B$ an effacement, $u_B\circ f:A\hookrightarrow M_B$ is also an effacement, making $$\begin{array}[ccccc]{} 0&\to&A&\stackrel f\to&B\\&&\|&&\downarrow u_B\\0&\to&A&\stackrel{u_B\circ f}\to&M_B\end{array}$$ commutative. This diagram extends to a morphism between the exact sequences. Passing to long exact sequences, we can decompose connecting maps of the arbitrary exact sequence by maps of effacement exact sequence, which we've already worked out. $\endgroup$ – Cave Johnson Oct 14 '18 at 14:12
  • $\begingroup$ @CaveJohnson Thanks a lot, I understand your solution. $\endgroup$ – user219967 May 20 '19 at 11:58
1
$\begingroup$

I have not read your whole question, but does [Weibel], Theorem 2.4.7 help you? (There for left derived functors and enough projectives, but this can be easily modified.)

$\endgroup$
  • $\begingroup$ Does that method still work if the category $\mathscr{A}$ doesn't have enough injectives? $\endgroup$ – D. Wynter Jan 25 '17 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.