Let $N$ and $M$ be a manifolds of respectively dimensions $n$ and $m$. If a smooth map ( $M$ from $N$ )is an immersion at a point $p$ in $N$ then it has constant rank $n$ in a neighborhood of $p$. If a smooth map is a submersion at a point $p$ in $N$ then it has constant rank $m$ in a neighborhood of $p$.
I want to prove this theorem but I could not. Can you help me.
This does not require the constant rank theorem. Some suggestions:
Since the matter is local, we can just as well work with open subsets $U\subset \mathbb R^n$ and $V\subset \mathbb R^m$, and a smooth map $f:U\to V$.
Being either an immersion or submersion at $p$ means that the differential matrix $Df$ has maximal rank at $p$. (Maximal rank of a matrix of size $m\times n$ is $\min(m,n)$.)
In general, the rank of a matrix is lower semicontinuous function of the matrix. This is because if the rank is $r$, the matrix has an $r\times r$ submatrix with nonzero determinant. By continuity, that determinant will be nonzero for all nearby matrices; therefore their rank is at least $r$.
Combine 2 and 3 to conclude that the set of matrices of maximal rank is an open subset of the set of $m\times n$ matrices.
Linear independence is an open condition: if $v_1,v_2,\ldots,v_k$ are a linearly independent set of vectors, then you can wiggle them a little bit, and they'll still be linearly independent.
For immersion: $i:M\rightarrow N$, you only need to worry about rank(d$(i)_x$) $ < m$ (since rank(d$(i)_x)\leq m$. But you know a) d$(i)_x$($v$) vary continuously and b) rank(d$(i)_p$) = m. Since linear independence is an open condition, there is a neighborhood around $p$ so that for all $x$ in this neighborhood rank(d$(i)_x)$ =rank (d$(i)_p$) = $m$.
For submersion: $p:M\rightarrow N$, you only need to worry about rank(d$(p)_x$)$< n$ (since rank(d$(p)_x)\leq n$. Now apply same argument.
