The difference between f(1,g) and f(g,1) is generally an issue of whether mathematicians give preference to "domains" or "ranges" of maps.
Here is one way that you could think of this. I can write EG for a category whose objects are objects are elements of G, and where each pair of objects has a unique map between them. This category has an action of G on it, and you can ask about G-equivariant functors from this category to another category what has a G-action on it.
To define such a functor on the level of objects it suffices to define F(1), where 1 is the unit; equivariance forces us to define F(g) = g F(1). On the level of morphisms, however, we have to make a choice. The unique morphism g→h becomes a morphism F(g)→F(h), and to make such maps compatible with the G-action it suffices to make one of the following sets of choices:
- We could define maps fh:F(1)→F(h) for all h, and get all the other maps as g fh:F(g)→F(gh). To be a functor, we need this to satisfy the cocycle condition fgh = (g fh) fg.
- We could define maps dh:F(h)→F(1) for all h, and get all the other maps as g dh:F(gh)→F(g). To be a functor, we need this to satisfy the cocycle condition dgh = dg (g dh).
In group cohomology, H1(G,M) classifies splittings in the semidirect product of G with M, and the cocycle condition we get comes from our convention of writing this group as pairs (m,g) (which is in the same order as the exact sequence it fits into) and not (g,m). Similarly for H2(G,M).
I would say that I've hit nonstandard cocycle definitions several times because I've been too lazy to come up with sensible conventions about when I'm thinking about domains and ranges or trying to sweep it under the rug, especially when dealing with Hopf algebroids and cohomological calculations there.
I don't have a good answer for higher cocycle conditions other than saying that writing 2-cochains using f(g,1,h) is somehow more unusual than either of the other 2 choices because it's somehow derived from focusing on the "middle" object in a double composite of maps.
To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{*}(\mathbb{Z}) = \oplus_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that
$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$
is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely
$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$
and one by applying $\Omega^{-1}\mathbb{Z} \otimes_\mathbb{Z}$
$\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} a \stackrel{1_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} \Omega^{-1}a$
The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.
To be completely explicit about this there are two functors naturally isomorphic to $\Omega^{-1}$ namely $\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z}(-)$ and $\mathbb{Z}\otimes \Omega^{-1}(-)$ since $\otimes_\mathbb{Z}$ is biexact there are natural transformations commuting $\Omega^{-1}$ with $\otimes_\mathbb{Z}$ namely
$\Omega^{-1}(-)\otimes_\mathbb{Z} (-) \stackrel{\sim}{\rightarrow} \Omega^{-1}((-)\otimes_\mathbb{Z}(-)) \stackrel{\sim}{\leftarrow} (-)\otimes_\mathbb{Z} \Omega^{-1}(-)$
Our example triangles above which can be obtained from one another by first moving the loops to the right and then applying the unit transformation show that there must be a sign attached to this map for this to give an isomorphism of these triangles. In particular for $\otimes_\mathbb{Z}$ to be compatible with the triangulated structure the two natural isomorphisms moving $\Omega$ about must have different signs. These natural isomorphisms are the precise cause of the sign change.
Best Answer
Topologically, you could say that this is true because $K(A,1)$ exists for nonabelian groups $A$. When the action of $G$ on $A$ is trivial, at least, $H^1(G,A)$ should be homotopy classes of maps from $K(G,1)$ to $K(A,1)$ (the same way $H^n(G,A) = H^n(K(G,1);A) =$ homotopy classes of maps $K(G,1) \to K(A,n)$ for $A$ abelian). In a similar way, $H^0$ is defined with coefficients in any pointed set.