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Is the following true?

Suppose $A\subset X$ is closed and $H\colon X\to Y$ is a homeomorphism.

Then $\partial H(A) = H(\partial A)$.

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up vote 2 down vote accepted

Since $H$ is bijective and $A$ is closed $$H(\partial A)=H(\overline A\setminus \operatorname{int}(A)) =H(A\setminus \operatorname{int}(A))=H(A)\setminus H(\operatorname{int}(A))$$ and $H$ maps closed sets to closed one $$\partial H(A)=\overline{H(A)}\setminus\operatorname{int}(H(A)=H(A)\setminus\operatorname{int}(H(A)),$$ we have to show that $H(\operatorname{int}(A))=\operatorname{int}(H(A))$. Note that since $H(\operatorname{int}(A))\subset H(A)$ and $H(\operatorname{int}(A))$ is open, we have $H(\operatorname{int}(A))\subset \operatorname{int}(H(A))$. Now, let $x\in \operatorname{int}(H(A))$. Then $x$ is in an open set $O$ which is contained in $H(A)$. Since $H^{-1}(x)$ is in $H^{-1}(O)$, open set contained in $A$, $H^{-1}(x)$ is in $\operatorname{int}(A)$, and $x=H(H^{-1}(x))$ is in $H(\operatorname{int}(A))$.

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