In my opinion, all answers go a little too far. In this (non-topological!) setting I think about this as follows: The aim is to analyze the lack of right-exactness of a left-exact functor $\Gamma:C \rightarrow D$ between abelian categories. One functorial notion that captures this inexactness is a delta-functor $\lbrace H^n, n \in \mathbb{N}$ with $H^0 = \Gamma$. This is what I call a cohomology theory for $\Gamma$. Now, there may exist a lot of them, and so it makes sense to look for universal cohomology theories for $\Gamma$, where universal means, that any morphism in degree 0 already uniquely extends to a morphism of delta-functors. Up to canonical isomorphism there exists only one universal cohomology theory for $\Gamma$ which is then called the right derivative of $\Gamma$. The question is of course if such a universal cohomology theory for $\Gamma$ exists. And here it comes: If the category $C$ has enough injectives, then $\Gamma$ has a right derivative which can be computed by injective resolutions. (You can confer Lang's Algebra book for all the above notions, it's actually pretty nice).

If you now want to get rid of all choices, you can move to the derived category.

Hope this helps.

In my opinion, all answers go a little too far. In this (non-topological!) setting I think about this as follows: The aim is to analyze the lack of right-exactness of a left-exact functor $\Gamma:C \rightarrow D$ between abelian categories. One functorial notion that captures this inexactness is a delta-functor $H = \lbrace H^n, n \in \mathbb{N}$ with $H^0 = \Gamma$. This is what I call a cohomology theory for $\Gamma$. Now, there may exist a lot of them, and so it makes sense to look for universal cohomology theories for $\Gamma$, where universal means, that given any delta functor $K = \lbrace K^n, n \in \mathbb{N} \rbrace$ (not necessarily with $K^0 = \Gamma$ !) and a morphism $f^0:H^0 \rightarrow K^0$ in degree zero, then there exists a unique extension of $f^0$ to a morphism $f:H \rightarrow K$ of delta-functors. Up to canonical isomorphism there exists only one universal cohomology theory for $\Gamma$ which is then called the right derivative of $\Gamma$. The question is of course if such a universal cohomology theory for $\Gamma$ exists. And here it comes: If the category $C$ has enough injectives, then $\Gamma$ has a right derivative which can be computed by injective resolutions. (You can confer Lang's Algebra book for all the above notions, it's actually pretty nice).

The universality condition forces a strong connection between $\Gamma$ and $H$. This allows you for example to extend constructions in degree 0 to arbitrary degree. An example from group cohomology: Let U be a subgroup of a group G and let $g \in G$. Then conjugation is an isomorphism of functors $(-)^U \rightarrow (-)^{gUg^{-1}}$, where $(-)^U$ is the U-invariant functor on G-modules. Now, $H^n(U,-)$ is a universal cohomology theory for $(-)^U$ and $H^n(gUg^{-1},-)$ is a universal cohomology theory for $(-)^{gUg^{-1}}$. Hence, the exists a unique isomorphism of delta-functors $H^n(U,-) \rightarrow H^n(gUg^{-1},-)$ and in this way we have extended the conjugation to all cohomology groups. This is very inexplicit of course, but this is a powerful property for more advanced constructions. And verifying each time that constructions are compatible with the deltas is not a nice job (normally people simply say, it works).

In Lang's book "Topics in the Cohomology of Groups" you will find a lot more of such arguments.

If you now want to get rid of all choices, you can move to the derived category.

Hope this helps.