Compute the Derived functor $RF^i(A)$ for $i\ge 1$

Question

Let $\mathscr{A}$ be a category with enough injective,$F$ be the left exact functor ,we can compute the right derived functor as follows:

First since enough injective,we can construct a injective resolution of $A$,as $0\to A\to I^{\cdot}$ , with $I^{\cdot} $ starting from zero degree.

I have three questions that confuse me :

1. If we compute the cohomology $H^{i}(I^{\cdot}) = 0$ for $i\ge 1$ correct?since the resolution is a exact sequence
2. To compute the right derived functor $RF^{i}(A)$ we need to take $F$ on the resolution,then we get the exact sequence $0\to F(A)\to F(I^{\cdot})$,then why the $RF^{i}(A)$ may not equal to zero for $i\ge 1$?
3. Why $RF^{i}(A) = 0$ for $i\ge 1$ when $F$ is not only left exact but exact?

Answer 1

After several months, I can answer this question now:

This question is in fact a bit subtle since we have $I^\cdot$ a complex of an injective object, for any left exact functor $F$ we have that injective implies $F-$injective, therefore if $I^\cdot$ is exact you have $F(I^\cdot)$ also exact. However if $$0\to X\to I^\cdot$$ exact it does not implies the exactness of $$0\to F(X)\to F(I^\cdot)$$

Look at the diagram below, if we want to prove exactness, we can decompose the long complex to short complex as follows:

Since $X$ is not injective, the sequence $$0\to X\to I^0\to \ker d_I^0 $$ needs not to split. Therefore the functor $F$ can not preserve the exactness as spliting sequence did.

Similarily since it's not clear $\ker d_I^k$ for $k\ge 1$ is injective or not, the splitting condition does not holds, therefore it can not guarantee the exactness of the sequence anymore.