Can $f:M^n \rightarrow N^m$ be an immersion if $m

Question

Let $M,N$ be manifolds with $\dim M=n$ and $\dim N=m$.
We say $f\colon M^n \rightarrow N^m$ is an immersion if:

$$df_p\colon T_pM \rightarrow T_{f(p)}N$$ is injective. Where $p \in M$.
Can there exist an immersion if $\dim M>\dim N$?

Answer 1

To add a bit of background to the comment of Matthew Leingang:

The dimension of the tangent spaces is that of the respective manifold due to construction. So we have $\dim T_pM=\dim M=m$ and $\dim T_{f(p)}N=\dim N=n$. According to the dimension formula (basic linear algebra, hence doesn't even need knowledge about manifolds), you have: $$ \dim\ker(\mathrm df_p) +\underbrace{\dim\operatorname{img}(\mathrm df_p)}_{\leq n} =\dim T_pM=m.$$ Here you can see, that if $n<m$ holds, then $\dim\ker(\mathrm df_p)\neq 0$, so the kernel can't be trivial, which is equivalent to $\mathrm df_p$ being injective.