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I got stuck at an exercise, can anybody help me? Let $f:X\to Y$ be a surjective map between two topological spaces satisfying the condition: for any subset $A$ of $X$, the interior of the image of A is contained in the image of the interior of $A$. Show that $f$ is continuous.

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Both for finding and for understanding this proof, it may help to keep in mind the simple example of a surjective function $\mathbb R\to\mathbb R$ with a single jump discontinuity.

Assume $f$ is not continuous. Then there is an open set $U$ such that $f^{-1}(U)$ is not open. Since $f$ is surjective, $f(f^{-1}(U))=U$, which is open. Since $f^{-1}(U)$ is not open, it is not equal to its interior. Thus, there is a point $x$ which is in $f^{-1}(U)$ but not in its interior. Let $y=f(x)$, and form a set $A$ by removing all preimages of $y$ except for $x$ from $f^{-1}(U)$. Since $x$ was not in the interior of $f^{-1}(U)$, it is not in the interior of $A$. Thus the image of the interior of $A$ does not contain $y$, since we removed all other preimages of $y$. However, $f(A)=f(f^{-1}(U))=U$, which is open, so the interior of the image of $A$ does contain $y$. The contradiction shows that $f$ must be continuous under the given assumptions.

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As a small addendum, a continuous surjection need not have this property: take $X=\{0,1\}$ with topology $\{\varnothing,\{0\},X\}$ $Y=\{0\}$, $A=\{1\}$, and $F$ the only map of $X$ onto $Y$. –  Brian M. Scott Sep 29 '11 at 8:46
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