From John M. Lee, Introduction to Topological Manifolds (2nd ed. 2011, pp. 20 & 27) (in 1st ed. 2000, page 18 & 21, the wording is similar, but the relevant propositions are differently numbered):
If X is a topological space and $p \in X,$ a neighborhood of $p$ is just an open subset of $X$ containing $p.$ [...] (In some books, the word "neighborhood” is used in the more general sense of a subset containing an open subset containing $p$ [...]; but for us neighborhoods are always open subsets.) [...]
In metric spaces, one usually first defines what it means for a map to be continuous at a point (see Appendix B), and then a continuous map is one that is continuous at every point. In topological spaces, continuity at a point is not such a useful concept. However, it is an important fact that continuity is a “local” property, in the sense that a map is continuous if and only if it is continuous in a neighborhood of every point. The precise statement is given in the following important proposition.
Proposition 2.19 (Local Criterion for Continuity). A map $f: X \to Y$ between topological spaces is continuous if and only if each point of $X$ has a neighborhood on which (the restriction of) $f$ is continuous.
It is easy to prove (in much the same way as Lee proves this lemma), that the same result is true with the word "open" in place of "continuous".
To be explicit (the result is so easy that it hardly seems necessary, but it makes for easier reference - also, even with "easy" results, it is equally easy to make silly mistakes - at least for me it is):
Lemma (Local Criterion for Openness) A map $f: X \to Y$ between topological spaces is open if and only if each point of $X$ has a neighbourhood on which (the restriction of) $f$ is open.
Proof. The "only if" part is trivial ($X$ itself is a neighbourhood of each of its points). For the "if" part, let $U$ be open in $X$. Any point of $f(U)$ is of the form $f(x)$, where $x \in U$. Given such $x$, let $V$ be a neighbourhood of $x$ on which $f$ (restricted to $V$) is open. Then $U \cap V$ and $f(U \cap V)$ are open, and $x \in U \cap V \subset U$, therefore $f(x) \in f(U \cap V) \subset f(U)$, so $f(x)$ has a neighbourhood contained in $f(U)$. Therefore $f(U)$ is open. Q.E.D.
This is stated as an exercise, for metric spaces only, in Arlen Brown & Carl Pearcy, An Introduction to Analysis (1995), but I haven't seen it stated anywhere else. (My search was not thorough.)
Is it not stated anywhere (in full generality) because, unlike the result for continuous functions, it is not an "important" result? Or, can anybody give a reference to its general statement? Does anybody know of any applications for it? (I ask out of idle curiosity only.)
