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Nov 16, 2018 at 13:10 comment added user120487 This seems to be correct by the usual (and trivial, sorry) argument of taking $f:= id: V_{\sigma} \rightarrow V_{\tau}$ and the symmetric variant of it.
Nov 16, 2018 at 12:27 comment added Jochen Wengenroth If the Mackey topology were the initial topology w.r.t. $V'$ it would be equal to the weak topology which (in many case, e.g., for infinite dimensional Banach spaces) is not the case.
Nov 16, 2018 at 11:56 history edited user120487 CC BY-SA 4.0
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Nov 16, 2018 at 11:55 comment added user120487 Thank you for your answer. I will edit my question: is the Mackey topology initial w.r.t. V' (in the above sense)?
Nov 16, 2018 at 8:10 history edited Jochen Wengenroth
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Nov 16, 2018 at 8:05 comment added Jochen Wengenroth Every topology $T$ on a set $X$ is initial, e.g. w.r.t. the identity $X \to (X,T)$. It is thus not clear what you are asking for.
Nov 15, 2018 at 21:00 review First posts
Nov 15, 2018 at 21:03
Nov 15, 2018 at 19:07 history asked user120487 CC BY-SA 4.0