The Stacks project

A silly question: how can we deduce that the pro-objects $( H^ p(E_ n) )$ and $( H^ p(D_ n) )$ are isomorphic?

Answer: because taking cohomology is a functor. Leaving as is for now.

Thanks for your answer. By taking cohomology, do you in fact extend the functor $H^ p( - )$ to $\mathrm{Pro}(D(\textit{Ab}))\rightarrow \textit{Ab}$ while require that the extension commutes with projective limit ?

Short answer: yes. Long answer: If $F : \mathcal{A} \to \mathcal{B}$ is a functor between categories, then $F$ determines a functor between pro-categories. Eg if you have an inverse system $(A_n)$ of $\mathcal{A}$, then you obtain an inverse system $(F(A_n))$ of $\mathcal{B}$. That functor sends isomorphisms (which are called pro-isomorphisms if we use the language of pro systems) to isomorphisms. Eg if $(A_n)$ and $(A'_n)$ are pro-isomorphic, then we obtain a family of maps $f_n : A_{\varphi(n)} \to A'_{n}$ and similarly $g_n : A'_{\psi(n)} \to A_n$ for some (rapidly growing) functions $\varphi, \psi : \mathbf{N} \to \mathbf{N}$ satisfying some relations between them and the transition morphisms of the systems. Then we just consider $F(f_n)$ and $F(g_n)$ in the category $\mathcal{B}$ and these morphisms satisfy the same relations in $\mathcal{B}$, hence the inverse systems $(F(A_n))$ and $(F(A'_n))$ are pro-isomorphic. You only ever apply $F$ to the objects and arrows of $\mathcal{A}$, so I think there cannot be any confusion.