I am reading the following lemma from Washington's book "Introduction to Cyclotomic Fields":
On the other hand, there is a counterexample, given by this answer. The comments below this answer indicate that for projective limits to be exact, one should add Hausdorff condition, whose necessity I don't really understand.
So now I get very confused because I have a counterexample and a proof, and I am not able to locate a potential flaw in the proof, which does not assume (at least explicitly) Hausdorff-ness.
I would appreciate it very much if anyone can help me understand the subtle issue here.
Presumably in the book you quote, "compact groups" are assumed to be Hausdorff (either because this is part of their definition of a topological group or because it is part of their definition of "compact"). The Hausdorff assumption is used when they conclude that $\phi_{i,i-1}^B(b_i)=b_{i-1}$ by continuity. All continuity gives you is that $\phi^B_{i,i-1}(b_i)$ is a limit of $(b_{i-1}^{(N)})$ (with respect to a subnet of the sequence along which $(b^{(N)})$ converges to $b$). The Hausdorff assumption then tells you that $b_{i-1}$ is the only such limit (since limits are unique in Hausdorff spaces), and so $\phi_{i,i-1}^B(b_i)$ must be equal to $b_{i-1}$.