I would like to give an explicit description of the first derived functor of familiar functors. Let me start with an example to explain what I need exactly.
Fix a group $G$ and consider the category of (left) $G$-modules. Let us compute the first right derived functor of the functor of invariants $\cdot^G$ (which is covariant and left exact), in the following manner.
Take an exact sequence $$0\to M'\to M\to M''\to 0$$ of $G$-modules (I will assume that $M'\subseteq M$ to simplify notation). Take $m''\in M''^G$ and ask when it comes from an invariant of $M$. To do this we lift $m''$ to an $m\in M$ chosen randomly. Now we would want $m-gm=0$ for every $g\in G$, so consider the function $G\to M'$ defined by $g\mapsto m_g:=m-gm$. The key point here is that since $m$ and $gm$ map to the same $m''$, the function is $M'$-valued. Now the standard step is to observe that the following "cocycle condition" is satisfied: $$m_g+gm_h=m_{gh}$$ and hence define $$Z^1(M',G):=\{a:G\to M'|\ a_g+ga_h=a_{gh}\}$$ ($a_g=a(g)$).
Question 1: why couldn't we simply define $$Z^1(M',G):=\{a:G\to M'\}?$$ I ask this, because with other functors I found hard if not impossible to find an appropriate cocycle condition.
One then proceeds to define the boundaries, but those I found straightforward so I won't explain them in detail. Turns out: $$B^1(M',G):=\{a:g\to M'|\ \exists m'\in M': a(g)=m'_g\}.$$
Question 2: if the answer to 1 is yes, then it could well be that the $H^1$ obtained as above (that is, take all the functions) and the standard one (take the functions satisfiying the cocycle condition) are the first strictly smaller than the second, so the first one can't be the right cohomology... where does it break? It seems to me that both satisfy the usual properties (functoriality, long exact sequence,...). This has to do with the universal $\delta$-functor stuff, but then how do I check that the first one is the right cohomology?
Let me try to carry this out in another example. Take the category of abelian groups (or modules over a ring) and consider the functor $\text{Hom}(P,\cdot)$ (so I am trying to define $\text{Ext}$). Again this is left exact and covariant. Start from $$0\to M'\to M\to M''\to 0.$$ If I take $f:P\to M''$, I can lift it to a function $\bar{f}:P\to M$. The condition for it to be a homomorphism is $$\bar{f}(p+q)-\bar{f}(p)-\bar{f}(q)=0$$ for any $p,q\in P$, so as before I construct the obstruction $F:P\times P\to M'$ given by $$F(p,q)=\bar{f}(p+q)-\bar{f}(p)-\bar{f}(q).$$ This is $M'$-valued for the usual reasons.
Question 3: what is the "cocycle condition" here? I couldn't come up with any.
On the other hand, the boundaries are again quite obvious to define.