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In my topology class, we have proved that if $f$ is a continuous function between two topological spaces, then $f$ preserves connectedness and it preserves compactness (assuming the domain is connected and/or compact).

We have also seen that $f$ preserves paths between two spaces.

What other structure does a continuous map preserve?

Is there some structure that a continuous map does not preserve?

user5826
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1 Answers1

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Suppose $f$ is a continuous map from the separable topological space $X$ to the topological space $Y$ and $A$ is a countable dense subset of $X$.

Let $V$ denote an open set in $Y$ with inverse image the open set $U$ in $X$. Let $a\in A\cap U$. Then $f(a)\in V$. Thus $f(A)$ is a countable dense subset of $Y$, so $Y$ is separable.

John Wayland Bales
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  • Is $f$ supposed to be surjective? Otherwise I don't understand the argument at all... – Najib Idrissi Nov 28 '16 at 09:08
  • No, I was not supposing $f$ to be surjective. Since a continuous map preserves compactness and connectedness only between the domain and the image, not between the domain and the codomain, I presumed that was the context of the question. If the domain of a continuous map is separable, so is the codomain. – John Wayland Bales Nov 28 '16 at 14:57
  • You mean that if the domain is separable, then so is the image (not the codomain), right? Besides in your answer you wrote "so $Y$ is separable" (it's easy to find counterexamples with $X$ separable, $f : X \to Y$ continuous, and $Y$ not separable, take $f$ to be constant for example), and you implicitly assumed that $U \neq \varnothing$ when you wrote "Let $a \in A \cap U$" (a dense set meets all nonempty open subsets...). – Najib Idrissi Nov 28 '16 at 15:03
  • You are right, I was implicitly assuming $f$ to be surjective since I was considering its image to be $Y$ regardless of whether or not $Y$ is a subspace of some "larger" space. The continuous image of a separable space is separable just as the continuous image of a connected space is connected and the continuous image of a compact space is compact. What has surjection to do with it? – John Wayland Bales Nov 28 '16 at 21:04