Let $f:M \rightarrow N$ be a smooth function between smooth manifolds $M,N$. A regular point is a point $x \in M$ s.t. $d_xf$ is surjective. Why does $\dim M> \dim N$ mean that there are no regular points in $M$?
I attempted to show this by contradiction: assume $x$ is regular,so $d_xf:T_xM \rightarrow T_{f(x)}N$ is a surjection. Thus $\forall y \in T_{f(x)}N,$ $ \exists$ $ v \in T_xM$ s.t. $d_xf(v) = y$. But dim $T_{f(x)}N=\dim N < \dim M = \dim T_xM$. But this is not a contradiction since we can still have a surjection from a higher dimension to one of lower dimension. This seems to show that actually $\dim M< \dim N$ means there are no regular points in $M$.
Where is this argument flawed, and what am I misunderstanding?
Maybe you wanted to say $\dim N>\dim M$. Intuitively the idea is that there is not enough space. But concretely, for a regular point the differential of $f$ must be surjective. Since the dimension of $N$ is bigger that the dimension of $M$ there is no way for a linear map to be surjective between the tangent spaces.