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Do we have

$f:(E,\tau)\to(F,\sigma)$ continuous if and only if $\forall A\subset E, f(\overline{A})= \overline{f(A)}$

or just: $f:(E,\tau)\to(F,\sigma)$ continuous if and only if $$\forall A\subset E, f(\overline{A})\subset \overline{f(A)}$$

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marked as duplicate by Henno Brandsma general-topology May 13 '18 at 20:48

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The later. For instance, take $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=\frac1{1+x^2}$ and $A=\mathbb R$. Then$$f\left(\overline A\right)=f(\mathbb{R})=(0,1],$$whereas $\overline{f(A)}=[0,1]$.

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  • $\begingroup$ And if $f$ is an homeomorphisem then we have the first one ? $\endgroup$ – Poline Sandra May 13 '18 at 18:21
  • $\begingroup$ @PolineSandra Of course. $\endgroup$ – José Carlos Santos May 13 '18 at 18:28
  • $\begingroup$ please why ? @José $\endgroup$ – Poline Sandra May 13 '18 at 18:39
  • $\begingroup$ @PolineSandra Because if $y\in\overline{f(A)}$, then $y=f(x)$ for some $x\in E$ (since $f$ is onto). If $x\notin\overline A$, then there is an open subset $O$ of $E$ such that $O\cap A=\emptyset$. Then $y=f(x)\in f(O)$ and $f(O)$ is an open set (since $f$ is a homeomorphism) which doesn't intersect $f(A)$ (since $f$ s injective). But then $y\notin\overline{f(A)}$, which is absurd. $\endgroup$ – José Carlos Santos May 13 '18 at 18:47

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