Exact sequence implies discrete topology

Question

I have been reading Goldman-Sah's paper about Locally Compact Rings of Special Types (https://core.ac.uk/download/pdf/82078567.pdf). On page 376, there is a statement: "Given an exact sequence $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$. If the topology of $M'$ is induced by the inclusion of its image in $M$, the topology of $M''$ is the factor topology of $M$, and finally that $M''$ is discrete". I was trying to prove it. If $K/M'$ subset of $M/M'$ with $M'\subseteq K$, then $K/M'$ is open if $K$ is open in $M$. If topology on $M$ is not discrete, does the topology on $M''$ also remain discrete? or I'm using the wrong set of $M/M'$? Thank you for any help.

Answer 1

This is how I read it:

The author only introduces the notion of a exact sequence that is "special for the relativisation" in the paragraph on 376. It's an algebraic exact sequence $0 \to M' \to M\to M'' \to 0$ such that

• $M$ has the topology induced by the injection $M' \to M$ (we know it's an injection from the algebra but we add a topological condition).
• $M''$ has the quotient topology wrt the canonical quotient with $M$ (we know algebraically that $M'' \simeq M{/}M'$ already, and we add the requirement that it's also topologically a quotient)
• $M''$ is discrete (just a very stringent condition, automatic when the quotient is finite, e.g.).

So it's quite a strict notion.. He then said he will not use this anyway, so personally I wouldn't dwell too long on this.