Let $(X,\tau)$ be a Hausdorff topological space. Let $Y=X/\sim$ a quotient topological space of $X$. Suppose that there exists a continuous function $f: Y \to X$ such that $\pi \circ f = Id_Y$ I am asked to show that $f(Y)$ is a closed subset of $X$. So far I've only been able to prove that $Y$ is also a Hausdorff space under this condition but I haven't been able to do much more.
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Let $x_1 \in X \setminus f(Y)$ and $x_2 = f(\pi(x_1))$. Since $X$ is Hausdorff, there are disjoint open neighbourhoods $U_i$ of $x_i$. Let $W$ be an open neighbourhood of $\pi(x_1)$ such that $f(W) \subset U_2$. Then
$$U_1 \cap \pi^{-1}(W)$$
is an open neighbourhood of $x_1$ not intersecting $f(Y)$.
answered Nov 6 '17 at 22:04
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$\begingroup$ And why does $U_1 \cap \pi ^{-1}(W) \cap f(Y) = \varnothing$? $\endgroup$ – McNuggets666 Nov 6 '17 at 22:49
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$\begingroup$ $W$ exists since $f$ is continuous. And $\pi^{-1}(W) \cap f(Y) = f(W)$. $\endgroup$ – Daniel Fischer♦ Nov 7 '17 at 13:34
