Since the definitions for a Hausdorff space and for a separated map is so similar, I wondered about the implications between them.

Suppose $X$ is a Hausdorff space, and we have a map $f:X \rightarrow Y$ to some topological space $Y$. So assuming that for two $x_1 \neq x_2 \in X$ we have that $f(x_1) = f(x_2)$, we know from the definition of a Hausdorff space that $f$ is separated.

Now onto the converse. Assuming that $f$ is separated. The Hausdorff condition on $X$ holds for all $x_1 \neq x_2 \in X$ when $f(x_1) = f(x_2)$, but since it's dependent on $f(x_1) = f(x_2)$ we cannot guarantee that the condition holds for all points $x_1 \neq x_2 \in X$. Is this correct?

Edit: Here's the definition of a separated map: A continuous map $f:X \rightarrow Y$ is called separated if any two points $x_1 \neq x_2 \in X$ with $f(x_1) = f(x_2)$ can be separated in the sense that there exists open sets $U_1, U_2 \subseteq X$ such that $x_1 \in U_1$, $x_2 \in U_2$ and $U_1 \cap U_2 = \emptyset$

  • $\begingroup$ Hello, what is the definition of a separapled map? $\endgroup$ – Paul May 20 '16 at 12:34
  • $\begingroup$ @Paul I edited my post to include the definition. $\endgroup$ – Auclair May 20 '16 at 12:45

The notion of separated map is the relative version of the notion of a Hausdorff space. By relative one means that it's a statement about the fibers of a map $f\colon X\rightarrow Y$.

A continuous map $f\colon X\rightarrow Y$ is separated iff all fibers $f^{-1}(y)$ are Hausdorff topological spaces for the induced topology on $f^{-1}(y)$. In particular, $X$ is Hausdorff iff the continuous map from $X$ into the one-point topological space is separated.

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  • $\begingroup$ I see. Thanks for the answer. Are the arguments in my post correct though? $\endgroup$ – Auclair May 20 '16 at 12:55
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    $\begingroup$ Yes. You proved that any continuous map defined on a Hausdorff space is separated, and you indicated that the existence of a separated map $f\colon X\rightarrow Y$ does not imply that $X$ is Hausdorff. Indeed, take any nonHausdorff $X$. Then the identity map on $X$ is separated. $\endgroup$ – Johannes Huisman May 20 '16 at 13:07
  • $\begingroup$ Fibers Hausdorff under subspace topology does not imply separated. Consider $f : \{a,b,c\} \to \{0,1\}, a \mapsto 0, b \mapsto 0, c \mapsto 1$ where open sets of $\{a,b,c\}$ are subsets containing $c$ and $\{0,1\}$ has the indiscrete topology. Then $f^{-1}(0) = \{a,b\}$ is Hausdorff under subspace topology, as is $f^{-1}(1)$. But $(a,b)$ is in the closure of the diagonal in $\{a,b,c\}\times_{f}\{a,b,c\}$ and not in the diagonal. Hausdorff should be replaced with relatively Hausdorff to fix this. $\endgroup$ – user577413 Apr 17 at 22:20

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