How are you proving it doesn't work if the path is quasi-linear? Can you prove that if f is quasi-linear, then the complement of f([a,b]) is dense-open? The hint seems to be suggesting the following strategy:
It was mostly my roommate's idea, it was some goofy thing that showed that M must then be globally homeomorphic to R^M. Or some such. We actually threw it out for something much more like what you suggested. We've made quite a bit of progress.
What we currently have is that we assume that f is onto and quasi-linear. Then, there exists a finite number, n, of U_i open in M such that h_i(U_i) is a homeomorphism, and the union of all these U_i is M, since f is onto. And also that, for each of these U_i, there exists [a_i, a_i+1] such that f( [a_i, a_i+1]) :=S_i is contained in U_i, and h_i(S_i) is a linear function. Then, since h_i(S_i) is a closed set with empty interior contained inside of an open set, every point of h_i(S_i) is a boundary point(Hand-wavey!), and so we have that X-h_i(S_i) is open and is dense in X, where X is the union of all h_i(U_i)'s.
Now I'm just sort of staring at the intersection of all X-h_i(S_i). This should be dense open
in X right? So if it is, then what?
I'm kind of stuck now.
Actually, do I really need to do this by contradiction?
Edit: Looking at Skeptic's suggestion, I've got 1 & 2, now I just need 3.
Edit 2: So, I think maybe I got it. If f_i is the restriction of f to [a_i, a_i+1], then f_i is quasi-linear. Thus there exists some U_i and h_i s.t U_i is open in M and h_i is a homeomorphism with h_i(U_i) :=V_i open and convex, and h_i(f_i) := S_i is linear. Then, V_i - S_i is dense and open in V_i.Then, letting Y be the union of all U_i's, h_i^-1(V_i-S_i) := W_i is dense and open in U_i, and thus is dense and open in Y. Then, Y- (Union of all W_i) is the intersection of all Y-W_i, and thus is a finite intersection of open dense sets. And thus, it is an open dense set in Y, and is thus non-empty.