The answer to the precise version of the question (the one asked at the end) is no.
The map (x,y)mapsto(x,xy) maps mathbbR2 onto all of mathbbR2 except the positive and negative parts of the y-axis. Following this with zmapstoz3 (where z=x+iyinmathbbC) yields a polynomial surjection (x,y)mapsto(f1(x,y),f2(x,y)) such that the preimage of (0,0) is the y-axis. Define f(x,y,z)=(f1(x,y),f2(x,y),z), so f defines a surjection mathbbR3tomathbbR3. Let U:=mathbbR3−lbrace(0,y,z):yleezrbrace, so U is a topological ball. The value of f on each deleted point (0,y,z) is the same as its value at (0,ez+1,z)inU, so f|U is still surjective.
For r>0, if uinU and f(u)=(0,0,logr), then u=(0,y,logr) with y>r, so |u|>r. So f(UcapB(r)) does not even contain (0,0,logr), let alone B(repsilon) for large r (for fixed epsilon>0).
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