I saw the following question:
An open set $U\subset\mathbb{R}^n$ contains the closed origin-centered unit ball $B = \overline{B(0, 1)}$. If a $C^1$ mapping $f:U\rightarrow \mathbb{R}^n$ with rank $n$ obeys $\|f(x)-x\|<1/2$ for all $x\in U$ show that
(1). $\|f\|^2$ must attain its minima in the interior of $B$
(2). $f(p)=0$ for some $p\in B$
I don't know what rank means in this question. I googled rank of a $C^1$ map and this wikipedia page says its the rank of the derivative.
I am confused whether it is the rank of the image of $f$ (which is the notion of rank we learn in linear algebra) or is it the rank of the derivative. Please help.
The rank of $f$ at $x\in U$ is by definition $\textrm{rank}(\mathrm{d}_xf)$.
Saying that $f$ has rank $k$, means that for all $x\in U$, the rank of $f$ at $x$ is equal to $k$.
In your case, using inverse function theorem, it is equivalent to say that $f$ is local diffeomorphism around each point of $U$.
The image of $f$ needs not to be a vector space, nor a submanifold of $\mathbb{R}^n$.
Example. Let $n=1$, $U=\mathbb{R}$ and $f\colon x\mapsto x^2$, then $f(U)=[0,+\infty[$.