Let $f:X \rightarrow Y$ be a homeomorphism and let $A \subset X$. Then $f$ restricts to a homeomorphism $f|_A:A \rightarrow f(A)$ between the subspaces $A$ and $f(A)$.

I am going to omit the restriction symbol when showing continuity? Any harm doing this?

Let $U$ be an open set in $f(A)$ as a subspace of $Y$. Then $U=f(A) \cap V$ for some $V$ open in $Y$. So $f^{-1}(f(A) \cap V)=f^{-1}(f(A)) \cap f^{-1}(V)=A \cap f^{-1}(V)$ which is open in $A$.

Would $f^{-1}(f(A))=A$ since $f$ is bijective? I believe it would.So $f|_A$ is continuous.

Let $W$ be open in $A$ as a subspace of $X$. Then $W=A \cap Z$ for some $Z$ open in $X$. Then $(f^{-1})^{-1}(A \cap Z)=(f^{-1})^{-1}(A) \cap (f^{-1})^{-1}(Z)=f(A) \cap f(Z)$ which is open in $f(A)$ since $f$ is bijective. So $(f|_A)^{-1}$ is continuous.

Now I'm attempting to prove bijectivity of $f|_A$.Suppose $f|_A(x)= f|_A(y)$. Then since $f|_A(x)=f(x)=f(y)=f|_A(y)$ and $f$ is injective $x=y$.

Onto surjective. I don't know I figure it would be automatic. Or would it? Any help here? Wait isn't any injective function with codomain restricted to its range necessarily a bijection?


  • $\begingroup$ Your proof is fine, but it might be easier to observe that $f|_A=f\circ i$ where $i$ is inclusion of $A$ into $X$. $\endgroup$ Apr 4, 2021 at 2:01

1 Answer 1


I prefer not to go into details of subspace topologies etc. but rely on general facts about initial topologies (as outlined here e.g.).

To simplify notation call $f\restriction_A$ just $f_A$, once you've fixed $A$. Then $f_A: A \to Y$ and $f_A = f \circ i_A$ where $i_A:A \to X: i_A(x)=x$ is the canonical injection. The subspace topology on $A$ is chosen such that $i_A$ is continuous (and minimally so), so $f$ continuous implies $f_A$ continuous is immediate. Also $i_B: B \to Y$ where $B = f[A]$ is likewise a continuous map and using the universal property for initial topologies and $$i_B \circ f' = f_A = f \circ i_A$$

$f': A \to f[A]$ being the map we have to prove continuous, we see that $f'$ is continuous iff $f_A$ is continuous and we know the latter to be true.

So continuity of $f$ is no problem. You could do a siimailr argument for $g: Y \to X$ which is the inverse of $f$. Being bijective, $g$ maps $f[A]$ back to $A$ etc. No topology, just basic set theory.

So $g': f[B] \to A$ is also continuous when $g$ is (which it is), so $f'$ is a homeomorphism, as $g'$ is the inverse of $f'$ etc.

Going more general makes stuff easier sometimes.


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