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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?

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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.

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  • $\begingroup$ 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

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