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Eric claims in this answer that, for $X$ and $Y$ topological spaces, then the following are equivalent

  1. If for all continuous maps $c:\mathbb R \to X$, $f \circ c$ is continuous then $f:X \to Y$ is continuous
  2. A subset of $X$ is closed iff its preimage under every continuous map $c: \mathbb R \to X$ is closed.

I can see how (2) implies (1), but not the converse. What am I missing?

Fernando Chu
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  • I believe (1) should quantify over $Y$. Hint: let $Y$ be $X$ with the topology that $U$ is open in $Y$ iff for all continuous maps $c : \mathbb{R} \to X$, we have $c^{-1}(U)$ is open. Show that $Y$ and $X$ have the same topology. – Mark Saving Apr 13 '22 at 20:43

1 Answers1

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Here is a proof of $(1) \Rightarrow (2)$.

We have to show that $A \subset X$ is closed if for each continuous $c : \mathbb R \to X$ the preimage $c^{-1}(A)$ is closed in $\mathbb R$.

So let $A \subset X$ be a subset such that for each continuous $c : \mathbb R \to X$ the preimage $c^{-1}(A)$ is closed in $\mathbb R$. $\{ \emptyset, X, X \setminus A \}$ is a topology on the set $X$; call the resulting space $Y$. Its closed sets are $\emptyset, X, A$. Now consider the identity function $f : X \to Y, f(x) = x$. We have $f^{-1}(A) = A$.

We shall show that $f$ is continuous which implies that $A$ is closed in $X$.

Let $c : \mathbb R \to X$ be continuous. We have $(f \circ c)^{-1}(A) = c^{-1}(f^{-1}(A)) = c^{-1}(A)$ which is closed in $\mathbb R$ by assumption. Since $(f \circ c)^{-1}(\emptyset) = \emptyset$ and $(f \circ c)^{-1}(Y) = \mathbb R$, the function $f \circ c$ is continuous. It follows from $(1)$ that $f$ is continuous.

Paul Frost
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