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I'm taking a course in elementary algebraic geometry, but I seem to be lacking a topological background. The following result is often used:

Let $X,Y$ be topological Spaces, and $U_i$ for $i \in I$ an open cover for $X$ (the convention here is that $\bigcup_{i \in I} U_i = X$). Then, a function $f: X \to Y$ is continuous if and only if $f|_{U_i}: U_i \to Y$ is continuous for $\forall i$ where $U_i$ is endowed with the subspace topology.

How does one prove this? I've not seen this lemma before. (Apparently it's called 'local property of continuity'?)

Albert
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1 Answers1

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One direction is trivial. For the other direction, suppose that $f\upharpoonright U_i$ is continuous for each $i\in I$. Let $V$ be an open set in $Y$; to show that $f$ is continuous, we need to show that $f^{-1}[V]$ is open in $X$.

For each $i\in I$ let $W_i=(f\upharpoonright U_i)^{-1}[V]$; by hypothesis $W_i$ is open in $U_i$, and since $U_i$ is open in $X$, this means that $W_i$ is open in $X$. And clearly

$$f^{-1}[V]=\bigcup_{i\in I}W_i\,,$$

so $f^{-1}[V]$ is open in $X$.

Brian M. Scott
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