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Dec 1, 2014 at 12:59 history edited user9072 CC BY-SA 3.0
updated MJ that does not work anymore
Mar 19, 2010 at 20:02 comment added gowers I don't want to spoil anyone else's fun, but it's not giving too much away to say that it can also be done without the axiom of choice. In fact, when I've set this question I've tended to get about as many constructions as people who seriously attempted the question.
Mar 19, 2010 at 19:48 comment added Ady Such a "locally surjective" map $f$ may be obtained as follows. Let {$1$} $\cup$ { $b_{t};t\in\mathbb{R}$ } be a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, and let us define the $\mathbb{Q}$ - linear map $f:$ $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ by $f(1):=0$ , and $f\left(b_{t}\right)=t$ $\left(t\in\mathbb{R}\right)$.
Mar 19, 2010 at 18:41 comment added gowers Incidentally, it's a nice exercise to find a map from the reals to the reals that takes every value in every open interval. Using that it isn't hard to find a map of the kind I'm claiming exists.
Mar 19, 2010 at 18:37 history answered gowers CC BY-SA 2.5