I was reading Milnor's 1958 notes in differential topology and came across the following theorem
Theorem: Let $U$ be an open set in $\mathbb{R}^n$ and let $f;U \to \mathbb{R}^p$ be differentiable, where $p \geq 2n$. Given $\epsilon$ > $0$, there is a $p \times n$ matrix $A = (a^i_j)$ with each $|a^i_j| <$ $\epsilon$, such that $g(x) = f(x) + Ax$ is an immersion.
Proof: $Dg = Df+A$; we would like to choose $A$ in such a way that $Dg(x)$ has rank $n$ for all $x$. i.e. $A$ should be of the form $Q-Df$, where $Q$ has rank $n$.
Define $F_k:\mathcal{M}(p,n;k) \times U \to \mathcal{M}(p,n)$ by the equation $$F_k(Q,x) = Q-Df(x)$$ $\mathcal{M}(p,n;k)$ is a submanifold of $\mathcal{M}(p,n)$ of dimension $k(p+n-k)$. $F_k$ is differentiable, and the domain of $F_k$ has dimension $k(p+n-k)+n$. As long as $k<n$, this expression is monotonic in $k$. Hence the domain of $F_k$ has dimension not greater than $$(n-1)(p+n-(n-1))+n=(2n-p)+pn-1$$ for $k<n$. Since $p\geq 2n$, this dimension is strictly less than $pn=$ dim($\mathcal{M}(p,n)$). Hence the image of $F_k$ has measure zero in $\mathcal{M}(p,n)$, so that there is an element $A$ of $\mathcal{M}(p,n)$, arbitrarily close to the zero matrix, which is not in the image of $F_k$ for $k=0, \dots$ , $n-1$. Then $A + Df(x)$ has rank $n$ for each $x$.
I dont understand the last part of the proof. i.e. How he concludes that $A+Df$ has rank $n$? Is it because $F_k$ is surjective and hence the chosen $A$ arbitrarily close to the origin satisfies $A=Q-Df(x)$ for some $x$ and $Q$ of rank $n$ and hence $A+Df(x)=Q$ has rank $n$ everywhere ? Please help. Thanks.
If $A+Df(x)=Q$ has rank $k<n$ for some $x \in U$, then $A=Q-Df(x)$ so that $A$ is in the image of $F_k$. This contradicts the fact that $A$ was chosen to be outside the image of $F_k$ for all $k<n$.