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Let $R$ be a topological ring, let $M$ be a topological commutative group, and let $R\rightarrow \operatorname{End}(M)$ be a ring homomorphism giving $M$ the structure of a left $R$-module (throughout, $\operatorname{End}$ will refer to endomorphisms in the category of topological groups).

The objective is to find a topology on $\operatorname{End}(M)$ that has the property that $R\rightarrow \operatorname{End}(M)$ is continuous iff the map $R\times M\rightarrow M$ is continuous.

My first attempt was:

Declare a net $\lambda \mapsto T_{\lambda}\in \operatorname{End}(M)$ to converge to $T\in \operatorname{End}(M)$ iff for every net $\mu \mapsto v_{\mu}\in V$ converging to $v\in V$, it follows that the net $(\lambda ,\mu )\mapsto T_{\lambda}(v_{\mu})$ converges to $T(v)$.

If this defines a topology, then I believe the topology will have the desired property and even make $\operatorname{End}(M)$ into a topological ring itself. The question is:

Does this definition of convergence define a topology? If not, what is a counter-example? If not this topology, is there any such topology? If so, what is it?

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    See this: http://mathoverflow.net/a/19288/64302. I think it shouldn't be hard to either verify each of the axiom, or get stuck and come up with a counter example2017-01-04
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    @user2520938 It's the axiom on iterated limits I had trouble with. This most certainly does suggest how to find a counter-example, but I was unsuccessful. That said, I spent much more time trying to prove it than find a counter-example. Perhaps I will try again tomorrow.2017-01-04

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