Let $(X,\tau_x)$ and $(Y,\tau_y)$ be topological spaces and $f:X \rightarrow Y$ be a homeomorphism. Prove that, if $\mathcal{B}$ is a basis of $$\tau_x = \{f^{-1}(O);O \in \tau_y\}$$ then $$B = \{f(B); B \in \mathcal{B}\}$$ is a basis of $\tau_y$.

My first step on this question was to prove that $\tau_x$ is a topology in $X$. Now, i don't really get how i can work with this basis. Any hint?


  • $\begingroup$ What have you tried? What are your thoughts about how being a homeomorphism relates to properties of images and preimages of open sets (wrt the respective topologies on target/domain)? $\endgroup$
    – J. Becker
    Mar 27, 2021 at 21:53
  • $\begingroup$ My attempt was to use the basis elements on the function (i think that's the basis) but it's kinda confusing to me. I tryed to make a parallel with basis on linear algebra and linear transformations. $\endgroup$ Mar 27, 2021 at 22:01
  • $\begingroup$ I'm not sure what you mean by "using the basis elements on the function". My immediate impression is that you probably aren't too familiar with the basic notions of topology, yet. You should probably recheck what the definitions mean precisely and make sure you understand well what exactly it is that you want to prove (i.e., what it means for B to be a basis of the topology on Y) and what it means to be a homeomorphism. :) $\endgroup$
    – J. Becker
    Mar 27, 2021 at 22:16
  • $\begingroup$ The notion of a basis is not clear to me, it looks weird to me that this function has to be a homeomorphism . $\endgroup$ Mar 27, 2021 at 22:30
  • $\begingroup$ Don't use an analogy with vector spaces, topological bases are quite different: it's a collection of open sets (so not of elements, as in a vector space) so that every open set is a union of some subfamily of the base. The main difference being that this union need not be at all unique in any way. There is no unique size of a base that could be called a "dimension" as in vector spaces. Though the minimal infinite cardinality of a base is a well-studied concept and called the weight of $X$. $\endgroup$ Mar 28, 2021 at 9:02

1 Answer 1


The fact that $(X,\tau_X) \to (Y,\tau_Y)$ is a homeomorphism implies that indeed $$\tau_X = \{f^{-1}[O]: O \in \tau_Y\}$$

the right to left inclusion is the statement that $f$ is continuous, and if $O \in \tau_X$ is open, $f[O]$ is in $\tau_Y$ (a homeomorphism is open, continuous and a bijection). and $f$ a bijection then implies that $O= f^{-1}[f[O]]$ which makes $O$ a member of the right hand set.

If $\mathcal{B}$ is a base for $\tau_X$, then $\{f[B]: B \in \mathcal{B}\}$ is a collection of open sets as $f$ is open and if $O \in \tau_Y$, $f^{-1}[O] \in \tau_X$ so $f^{-1}[O] = \bigcup \mathcal{B}'$ for some $\mathcal{B}' \subseteq \mathcal{B}$ because we have a base, and then

$$O = f[f^{-1}[O]]= f[\bigcup \mathcal{B}'] = \bigcup \{f[B]: B \in \mathcal{B}'\}$$ holds and so we can write any open $O$ in $Y$ as a union of $\{f[B]\mid B \in \mathcal{B}\}$, QED.

  • $\begingroup$ Get it! Thank you. $\endgroup$ Mar 28, 2021 at 18:02

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