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Let $g : X \to Z$ be a surjective continuous map. Let $X^* = \{g^{-1}(z) ~|~ z \in Z\}$, and give $X^*$ the quotient topology. (a) The map $g$ induces a bijective continuous map $f : X^* \to Z$ such that $f \circ p = g$, where $p : X \to X^*$ is the projection induces the quotient topology on $X^*$; $f$ is a homeomorphism if and only if $g$ is a quotient map. (b) If $Z$ Hausdorff, then $X^*$ is Hausdorff.

As the title suggests, I am working on part (b). Let $g^{-1}(z)$ and $g^{-1}(z')$ be distinct in $X^*$. Then $z \neq z'$ otherwise they won't be distinct. Since $f$ is bijective, $f(g^{-1}(z))$ and $f^{-1}(g^{-1}(z'))$ will be distinct in the Hausdorff space $Z$, which means there are disjoint open sets $U$ and $V$ in $Z$ containing $f(g^{-1}(z))$ and $f(g^{-1}(z'))$, respectively. Note that $\emptyset = U \cap V$ implies $\emptyset = f^{-1}(\emptyset) = f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(U)$, so that $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open sets. Note that the inverse image under $f$ is but the image under $f^{-1}$, so that $f^{-1}(f(g^{-1}(z)) \in f^{-1}(U)$ or $g^{-1}(z) \in f^{-1}(U)$, and similarly $g^{-1}(z') \in f^{-1}(V)$.

How does this sound? Seems a little roundabout taking the image and then the preimage, but that's the only proof I could think of.

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Hint: If $f\colon X\to Y$ is continuous and injective, while $Y$ is Hausdorff, then $X$ is Hausdorff; this is because preimages of disjoint open sets are disjoint and open.

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  • $\begingroup$ That is actually what OP proved for this special case. $\endgroup$ – amsmath Oct 12 '17 at 21:32
  • $\begingroup$ @amsmath: My suggestion was that it is easier to see the proof in the general case (without all the extra distracting information floating around). $\endgroup$ – tomasz Oct 15 '17 at 3:34
  • $\begingroup$ I see. This makes sense. $\endgroup$ – amsmath Oct 15 '17 at 4:05

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