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I am puzzled with the following statement: "Given any map $f:X\to Y$ where $X$ is equiped with the trivial topology $(\varnothing,X)$, then this map is continuous iff $Y$ has the trivial topology. (This can be found M.Gemignani - Elementary Topology, p70, §4.3, Example 8,). When the topology on $Y$ is not trivial, the author says that "the preimage by any map $f$ of a non-trivial open set (which exists) is not trivial (nor $\varnothing$ nor $X$)...hence the preimage is not an open subset, therefor $f$ could not be continuous".

Consider the constant map example: say $f(x)=a$, then $f^{-1}(U)=X$ where $U$ in any open set containing $a$, where as $f^{-1}(U)=\varnothing$ if $a\not\in U$ and we have checked that the preimage of any open set is open in $X$ since $X$ has the trivial topology. Then, any topology on $Y$ - trivial or not - makes the constant map continuous.

Where is the mistake?

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Pay attention he said the function f is "onto"!

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    $\begingroup$ Thanks for your answer, now this makes sense. "Given any surjective map $f:X→Y$ where $X$ is equiped with the trivial topology (∅,X), then this map is continuous iff $Y$ has the trivial topology". $\endgroup$ – NevD Sep 19 '14 at 14:03
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    $\begingroup$ Indeed, and the text does say onto. Which is a common synonym for being surjective. $\endgroup$ – Henno Brandsma Sep 19 '14 at 20:23

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