Assume, we have sets $A,B$ such that $A\simeq B$, i.e. there is a bijection $f\colon A\to B$.

Next, assume that we have equiped $B$ with some topology $\tau_B$ and have another topological space $(C,\tau_C)$ such that some function $h\colon B\to C$ is continuous with respect to to $\tau_B$ and $\tau_C$.

Does it then make any sense to consider $h$ as a continuous function from $A$ to $C$ since $h\colon (A\simeq~) B\to C$ and if yes with respect to which topology $\tau_A$ on $A$ is it continuous?


You can define a topology on $A$ by saying that $U \subset A$ is open if and only if $f(U)$ is open in $B$.

This gives a topology on $A$ such that $h \circ f : A \to C$ is continuous.

| cite | improve this answer | |
  • $\begingroup$ This means that $A$ and $B$, as topological spaces, are isomorph, or? $\endgroup$ – John_Doe Mar 31 '17 at 14:17
  • $\begingroup$ Yes, because they were isomorphic as sets and then we just copied the topological structure of $B$ to $A$ via $f$. $\endgroup$ – Elvorfirilmathredia Mar 31 '17 at 14:22
  • $\begingroup$ For topologically identical spaces we usually use the term "homeomorphic", not "isomorphic". $\endgroup$ – MartianInvader Mar 31 '17 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.