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I am trying to prove the following:

An Introduction to Homological Algebra - C. A. Weibel (1994)

Exercise 2.1.2) If $F~:A\to B$ is an exact functor, show that ${{T}_{0}}=F$ and ${{T}_{n}}=0$ for $n\ne 0$ defines a universal $\delta $-functor (of both homological and cohomological type )

Can someone teach me how to prove this?

Najib Idrissi
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math-ha01
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1 Answers1

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There is two things to prove :

First that $T$ is indeed a $\delta$-functor. For this, you only need to say that, because $F$ is exact, every short exact sequence in $A$ $$0\longrightarrow M'\longrightarrow M\longrightarrow M''\longrightarrow 0$$ gives a long exact sequence in $B$ : $$...\longrightarrow T_1(M')\longrightarrow T_1(M)\longrightarrow T_1(M'')\longrightarrow T_0(M')\longrightarrow T_0(M)\longrightarrow T_0(M'')\longrightarrow 0.$$

Then you need to prove that $T$ is universal. For this, you need to prove that to give a morphism of $\delta$-functors $T\rightarrow S$ is equivalent to give a natural transformation $T_0\rightarrow S_0$. But this is clear, because there is not much choice for the morphisms $T_n\rightarrow S_n$ if $n\neq 0$ : it must be 0 since $T_n=0$. The natural transformation $T_0\rightarrow S_0$ extends then uniquely as a morphism of $\delta$-functors.

Roland
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