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Let $X$ and $Y$ be topological spaces and suppose $f: X \to Y$ is continuous. If $f$ is continuous on $U \subset X$, will the restriction $f_U :U \to Y$ be continuous, if we consider $U$ to be a topological space of its own?

My second question is given open sets $U, V \subset \mathbb{R^n}$ and continuous functions $f_1 : U \to \mathbb{R^n}$ and $f_2 : V \to \mathbb{R^n}$ Will the function $f_{U \cup V}: U \cup V \to \mathbb{R^n}$ defined in the obvious way be continuous?

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1) Yes, if $U$ has the subspace topology. The preimage of an open set $V$ in $Y$ under $f_U$ is just $f_U^{-1}(V)=f^{-1}(V) \cap U$, which is open by the definition of the subspace topology on $U$. This is essentially why the subspace topology is defined the way it is, so that restrictions of continuous maps are continuous.

2) For such a function to be defined, we need $f_1$ and $f_2$ to agree on $U \cap V$. In this case, the function is continuous. The preimage of an open set under $f_{U \cup V}$ is just the union of the preimages of that set under $f_1$ and $f_2$. There's nothing particular about $\mathbb{R}^n$ here. See the Pasting lemma for more general conditions.

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    Yes. And since that’s the case, we know that one such $D$ is $f^{-1}[U]\cap X$ itself.2012-07-11