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The following is related to this post.

Let $X=X'$, as sets, and let $T:X \rightarrow X'$ be a surjective map from a countably infinite-dimensional LCS $X$ to itself and equip $X'$ with the final topology. Thus, $T$ is a quotient map.

When/(it possible) for the topology of $X'$ to be finer than that of $X$ or contain open sets not in the topology of $X$?

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  • $\begingroup$ Yes. Take a metrizable lcs $X_1$ and a non-metrizable lcs $X_2$. Then the projection operator $X_1\times X_2\to X_1$ is not a homeomorphism. $\endgroup$
    – Taras Banakh
    Commented Jun 23, 2020 at 5:09
  • $\begingroup$ @TarasBanakh Following your comment, I realized my initial formulation was trivial. I rephrased the question to better express what I had in mind. $\endgroup$
    – ABIM
    Commented Jun 23, 2020 at 8:17
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    $\begingroup$ Ok, then take a metrizable countably-dimensional lcs $X_1$ and the countably dimensional lcs $X_2$ with the strongest lcs topology (it is usually called the space $\varphi)$. Then $X_2$ is the image of $X_1\times X_2$ and algebraically, $X_2$ is isomorphic to $X_1\times X_2$. The space $X_2$ has the finer topology than $X_1\times X_2$ since the topology of $X_2$ is the largest lcs topology on $X_2$. $\endgroup$
    – Taras Banakh
    Commented Jun 23, 2020 at 8:31

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