Proof of Profinite groups are inverse limits of finite groups Suppose $G$ is a Profinite group so it is, by definition, inverse limit of finite groups endowed with the discrete topology. It is compact, Hausdorff and totally disconnected. $\mathfrak U $ be set of all open normal subgroups of $G$ then the canonical homomorphism $G \rightarrow \varprojlim_{N \in \mathfrak U} G/N $ is a homeomorphism.
So they are isomorphic. I am not sure if I understand the details of the proof clearly. Below, I have tried to expand on proof from https://www.math.ucla.edu/~sharifi/groupcoh.pdf#theorem.2.1.24
The canonical homomorphism $\phi$ has closed image (I do not understand where this is used ) as if $(g_N N)\notin \phi (G) $ then for some $N_1, N_2, \ g_{N_1} N_1 \neq g_{N_2}N_1 $. So the open set $$\pi _{N_1}^{-1}(g_{N_1}N_1) \cap \pi _{N_2}^{-1}(g_{N_2}N_2) \cap \varprojlim_{N \in \mathfrak U} G/N \ $$ is open nbhd around $(g_N N)$ and does not intersect with image of $G$.
The kernal is trivial as the space is totally disconnected and open normal subgroups form a basis of nbhds of $1$.
$(g_N N) $ is not in the image then the intersection $\cap _Ng_N N$ is empty as otherwise if $g \in \cap _N g_N N$ then $\phi (g)=(g_N N) $. So by finite intersection property $g_1N_1 \cap \cdots \cap g_k N_k=\emptyset $. If $1 \in M =\cap N_i $ then $g_M M \subset g_1N_1 \cap \cdots \cap g_k N_k =\emptyset  $ which is a contradiction. So the map is surjective.
Does it look good?
 A: To be honest, in the proof from the notes you link I do not see the need to observe that the map has closed image. There are some hints that perhaps these notes are a little rough. For instance:

*

*$Q$ is declared as a name for the inverse limit but never referenced again.


*They say "since $G$ is compact some finite subset of $\{g_N N\mid N\in\mathscr{U}\}$ is empty" when of course they mean that there are finitely many $g_{N_1}N_{1},\ldots g_{N_k}N_{k}$ from the set whose intersection is empty.
Anyway to show that this map $\phi$ is a homeomorphism this officially requires four things:

*

*$\phi$ is injective


*$\phi$ is surjective


*$\phi$ is continuous


*$\phi$ has continuous inverse
Your explanation of 1 and 2 looks good to me. For 3, the proof just declares that $\phi$ is continuous so this would be a good thing to double-check for your own understanding. Finally for 4, you can do this directly or use the fact that a continuous bijection from a compact space to a Hausdorff space is always a homeomorphism. See https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism
