# Why do universal $\delta$-functors annihilate injectives?

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Suppose $\mathcal{A}$ has enough injectives, and consider a universal (cohomological) $\delta$-functor $T^\bullet$ from $\mathcal{A}$ to $\mathcal{B}$. By the theory of derived functors, we know that that $T^n (A) = 0$ for all injective objects $A$ and all $n \ge 1$ and that $T^n$ is effaceable for $n \ge 1$. But can this be shown directly without invoking derived functors?

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If $T^*$ is universal you can show that $T^i$ must be the $i$th right derived functor of $T^0$ and so you get higher vanishing on injectives. This is because the $R^* T^0$ is also a universal delta functor, and hence (by universality) must be the same as $T^*$. In fact, universal delta functors are just a different formalization of derived functors, so I don't know why you'd want to avoid derived functors. –  Parsa Apr 2 '12 at 10:27
@Parsa: Universal $\delta$-functors make sense in the absence of enough injectives, but derived functors require some form of "enough acyclics" assumption. I am basically wondering if it is true that injectives are acyclic even when there are not enough injectives. –  Zhen Lin Apr 2 '12 at 10:33
Why do you want to show this without using derived functors? –  Jimmy R Jul 23 '13 at 21:18