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Let $X,Y$ be topological spaces and $f : X \rightarrow Y$ be a continous,surjective map.

Then which of the following are true -

1) if $X$ is separable then $Y$ is separable.

2) if $X$ is first countable then $Y$ is first countable.

3) if $X$ is Hausdorff then $Y$ is Hausdorff.

4) if $X$ is regular then $Y$ is regular.

From the comments Topological properties preserved by continuous maps , and the wikipedia article,I think the first option is correct as then $f(X) = Y$.

Can we add more properties to the bag, if we add some other conditions in addition to surjectivity of $f$ ?

BAYMAX
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1 Answers1

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You are right about the separability, but you should try to write down a proof.

Separation axioms are easily broken down this way. You can for example have a continuous surjection from $[0, 1]$ onto the two-point indiscrete space. So you start with something that is even metrizable compact and end with something that is not even $T_0$. And the map is even closed quotient.

You should think about what do you need from an equivalence to obtain the two-point indiscrete space as the quotient. Also, since you have asked about just sontinuous surjection, you can take identity and a coarser topology that fails to satisfy the separation axiom.

Regarding the first countability, you can start with any topological space $Y$ that is not first countable, and let $X$ be $Y$ with the discrete topology and $f$ again the identity. Even if you want $f$ to be quotient, this is not true. Just glue $ω$ copies of $[0, 1]$ at $0$. This is even a closed quotient. On the other hand, the claim is true if $f$ is open quotient or if $X$ is compact.

Other properties that are preseved by continuous srujections are compactness and conectedness, and the proofs are straightforward.

user87690
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