today in my topology class we had discussed about continuos map in topological spaces. And I have this question, for $(X,\tau)$ and $(X',\tau')$ two topological spaces and $f\colon X\to X'$, if $f(A)\subseteq X'$ is an open set $\forall A\in\tau$ It implies that $f$ is continuous? or what is necessary to get the continuity of the map?
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3The function you're describing is an "open map/function". Open maps need not be continuous. – Vercingetorix Dec 09 '20 at 00:52
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It is neither necessary nor sufficient for continuity. A function that satisfies this condition is called an “open map”.
For an example of a continuous function that does not satisfy this property, consider the real numbers with the usual topology, and a constant function $f\colon \mathbb{R}\to\mathbb{R}$. This is continuous, but the only open set whose direct image is open is the empty set; all other sets have image a singleton, which is not open.
For an example of an open mapping that is not continuous, consider $X=X’=\{a,b\}$, $\tau=\{\varnothing, X\}$, and $\tau’=\{\varnothing, \{a\}, \{b\}, X’\}$. Then the identity map is open, but not continuous. Other examples can be found here.
Arturo Magidin
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