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Exercise: Let $X$ and $Y$ be topological spaces, $f:X\to Y$ a continuous open map, and assume that for every $y\in Y$ the subset $f^{-1}\{y\}$ is irreducible (with respect to the subspace topology). Assume that $Y$ is irreducible. Show that $X$ is irreducible.

My answer: Assume $X$ is reducible, say with $U,V\subset X$ non-empty open sets such that $U\cap V=\emptyset$. Then $f(U)$ and $f(V)$ are open so $f(U)\cap f(V)\ne\emptyset$. Let $y\in f(U)\cap f(V)$. The subspace $f^{-1}\{y\}$ is irreducible and intersects both $U$ and $V$. Furthermore $U\cap f^{-1}\{y\}$ and $V\cap f^{-1}\{y\}$ are open sets in $f^{-1}\{y\}$, so $U\cap V\cap f^{-1}\{y\}\ne\emptyset$. Thus by contradiction $X$ is irreducible.

But I haven't used the fact that $f$ is continuous, so is that assumption superfluous, or have I made a mistake?

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Your proof is completely valid. It would seem that we can drop the assumption of continuity of $f$. That being said, continuity is often assumed as a matter of course.

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