Usually whenever one reads the definition of profinite group, one starts with an ordered set $I$ which is *directed*, meaning that for every $i,j\in I$ there is some $k\in I$ such that $i\leq k$ and $j\leq k$. Then one has a family of finite groups $\{G\_i\}\_{i\in i}$, compatible maps between them and defines $\varprojlim\_i G\_i$ as the subset of $\prod\_{i\in I}G\_i$ formed by coherent sequences.


The thing that is not clear at all to me is why we require $I$ to be directed, because it seems that dropping that assumption basically nothing would change and the definition would remain the same. Moreover, one thing we would like is for example to consider $\prod\_iG\_i$ a profinite group. This can be seen easily if we take the trivial ordering on $I$. But with this ordering, $I$ is not a directed set. Is there a way to see $\prod\_i G\_i$ as an inverse limit of finite group over a directed set? And more in general, if $\{P\_i\}\_{i\in I}$ is a family of profinite group and $P=\prod\_iP\_i$ is their direct product with the product topology, is $P$ always a profinite group?


Now let assume that $\{P\_i\}\_{i\in I}$ is a family of profinite groups for which the question above has an affirmative answer. Let $A$ be a discrete abelian group over which $P\_i$ acts trivially for all $i$'s. Is it true that


$$H^1\left(\prod\_iP\_i,A\right)\simeq \bigoplus\_iH^1(P\_i,A)$$
Here $H^1$ stands for the usual first cohomology group with coefficients in $A$.


Thanks a lot to anyone which is willing to give me a hint!



Profinite groups, directed sets and $H^1$
I hope this will clarify the situation:


I: Why is $I$ a directed set: In fact, there is no need for $I$ to be directed. One can define the inverse limit of a inverse system without the restriction that $I$ is directed. However, one should be aware that by doing so, certain propositions about profinite groups resp. spaces won't work anymore. 
For example the following proposition is not true anymore: Let $G=\varprojlim\_{i\in I}G\_i$ be a profinite group, where $I$ is a directed set and let $\varphi\_i:G\rightarrow G\_i$ be the projection homomorphisms. Then $\{S\_i:S\_i=\mathrm{ker}(\varphi\_i)\}$ is a fundamental system of open neighborhoods of the identity element of $1$ in $G$.


II: Is a direct product of profinite groups $\prod\_iP\_i$ (with the product topology) again profinite: Yes. See Proposition 2.2.1 in 'Ribes, Zalesskii, Profinite groups'.


III: Does $H^1\left(\prod\_iP\_i,A\right)\cong \bigoplus\_iH^1(P\_i,A)$ hold: Yes if $\prod\_iP\_i$ acts trivially on $A$. Set $P=\prod\_iP\_i$. For every $P\_i$ there is a collection of open-closed normal subgroups $\mathcal{N}\_i$ such that $H^1(P\_i,A)=\varinjlim\_{N\in \mathcal{N}\_i}H^1(P\_i/N,A)$ (See Corollary 6.5.6'Ribes, Zalesskii, Profinite groups'). Furthermore, there is collection of open-closed normal subgroups $\mathcal{N}$ such that $H^1(P,A)=\varinjlim\_{N\in \mathcal{N}}H^1(P/N,A)$. 
In fact, we can choose $\mathcal{N}\subseteq \prod\_i\mathcal{N}\_i$ to be the set of all $N\in \prod\_i\mathcal{N}\_i$ such that $M\_{N}:=\left\{k:\mathrm{pr}\_k(N)\neq P\_k\right\}$ is finite.
It is easy to see that we can embed each $\mathcal{N}\_i$ into $\mathcal{N}$ such that $\mathcal{N}\_i\cap \mathcal{N}\_j=\{P\}$ if $i\neq j$. We can rewrite $\varinjlim\_{N\in \mathcal{N}}H^1(P/N,A)$ as $$\varinjlim\_{N\in \mathcal{N}}H^1(P/N,A)=\left(\bigoplus\_{N\in \mathcal{N}}H^1(P/N,A)\right)/\mathcal{I}.$$ Each non-zero element of the subgroup $\bigoplus\_{i}H^1(P\_i,A)$ of $\bigoplus\_{N\in \mathcal{N}}H^1(P/N,A)$ is not contained in the subgroup $\mathcal{I}$, since $\mathcal{N}\_i\cap \mathcal{N}\_j=\{P\}$ for $i\neq j$. Hence $\bigoplus\_{i}H^1(P\_i,A)\subseteq H^1(P,A)$. It remains to verify $\bigoplus\_{i}H^1(P\_i,A)\supseteq H^1(P,A)$. Note that every $N\in \mathcal{N}$ can be written as a finite intersection $\bigcap N\_i$ of elements in $N\_i\in \mathcal{N}\_i$. Set $\tilde{\mathcal{N}}=\mathcal{N}\backslash \bigcup\_i\mathcal{N}\_i$. Then $H^1(P/\tilde{N},A)\cong \bigoplus\_{k\in M\_{\tilde{N}}}H^1(P\_k/\mathrm{pr}\_k(\tilde{N}),A)$. Therefore, it is easy to see that $H^1(P/\tilde{N},A)$ is contained in $\mathcal{I}$. Thus $\bigoplus\_{i}H^1(P\_i,A)=H^1(P,A)$. 


