Is a homeomorphism which maps lines to lines (and fixes zero) necessarily linear? We know the homeomorphism $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ maps straight lines to straight lines and the zero vector to the zero vector. Is it Linear?? If so, how can we prove it?
Are you asuming $n&gt;1$? If not a good counter-example could be the function $\phi(x)=x^3$ ( $x\in\mathbb{R}$) . It's a homeomorphism, maps $\mathbb{R}$ to $\mathbb{R}$ (the only line in $\mathbb{R}$) and obviously it maps $0$ to $0$
This is not enough to suggest linearity w/o it being a homeomorphism... It has satisfy $\phi(a*v_1 + b*v_2) = a\phi(v_1)+b\phi(v_2)$.  So lines to lines does not promise linearity, though it is a requirement as lines are a linear combination of vectors in $\mathbb{R}^n$.  Assuming you are looking at $\mathbb{R}^n$ as a combination of vectors in the typical sense.

The homeomorphism takes away the discontinuous cases and the situations with a kernel.  My bad I did not read the problem correctly initially.  

Lines can be expressed as a linear combination of elements in a vector space with an indeterminate to the first power.  Just express a matrix with such a basis as the mapping and you have you linearity in the form of a matrix.