I'm trying to solve the following question:

Let $F:M\longrightarrow N$ be a bijective map. Prove that, if $M$ is a topological space, then $N$ admits a unique topology making $F$ a homeomorphism.

I know that this topology can be expressed as $T_N=\{F(U):U\in T_M\}$, where $T_M$ is the topology defined on the space $M$, and I can prove that in this case $F$ is a homeomorphism between the topological spaces $(M,T_M)$ and $(N,T_N)$. But how can I prove the uniqueness of this topology?

  • $\begingroup$ Please, write your questions completely in english. $\endgroup$ Mar 4, 2017 at 14:27
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    $\begingroup$ Suppose $T_N$ is missing an $F(U)$ for some $U$ open in $M$. Would $F^{-1}$ be continuous? Suppose $T_N$ contains some $F(V)$ for some $V$ not open in $M$. Would $F$ be continuous? $\endgroup$
    – Kenny Wong
    Mar 4, 2017 at 14:27
  • $\begingroup$ @Daniel Robert-Nicoud: sorry, I didn't realize I hadn't written everything in English, the book where I found this problem was in Spanish and my brain somehow mixed languages. $\endgroup$ Mar 4, 2017 at 19:55

1 Answer 1


Suppose that $T$ is any topology that makes $F$ a homeomorphism. (which is an open, continuous bijection).

Then if $F[U] \in T_N$, for some $U \in T_M$, we know that as $F$ is open as a map from $(M,T_M)$ to $(N,T)$, that $F[U] \in T$, so $T_N \subseteq T$.

Remark: We can also use the continuity of $G = F^{-1}$ (the inverse function exists for bijections) and see in that way that $G^{-1}[U] = F[U]$ must be open in $T$ and prove this inclusion.

On the other hand if $U \in T$, $F^{-1}[U] \in T_M$ by continuity and so (by bijectiveness) $U = F[F^{-1}[U]] \in T_N$ by definition of $T_N$. So $T \subseteq T_N$.

So we have for any topology $T$ on $N$ that makes $F: (M, T_M) \rightarrow (N,T)$ a homeomorphism we have shown that $T = T_N$. So $T_N$ is unique: it fulfills the job, and there can be no other.

  • $\begingroup$ Thank you so much, this is perfect and so simple. But in your last paragraph, I think you mean "that makes F a homeomorphism" instead of "that makes F continuous", right? $\endgroup$ Mar 4, 2017 at 19:52
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    $\begingroup$ @WildFeather Indeed . I've edited it. $\endgroup$ Mar 4, 2017 at 22:46

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