I'm stucked in the following problem:
Let $f: \mathbb{R^m} \rightarrow \mathbb{R^n}$ be a differentiable map with $n \geq 2m$. Let $\varepsilon >0$, prove that there exists $\alpha : \mathbb{R^m} \rightarrow \mathbb{R^n}$ linear with $\Vert \alpha \Vert < \varepsilon$ such that $f+ \alpha$ is an immersion.
I suppose that the norm $\Vert \cdot \Vert$ does not play an important role, so take any matricial norm.
The first observation is that $T(f+ \alpha) = Tf + T\alpha = Tf + \alpha$, namely in each point $x \in \mathbb{R}^m$ one has $T_x(f+\alpha) = T_xf + \alpha$.
Next, $f$ being an immersion is by definition that for any $x \in \mathbb{R}^m$ the derivative $T_xf$ is of rank $m$. If $f$ is an immersion then take $\alpha$ to be the zero map. If it is not the case then $rank(T_xf)<m$ for some $x \in \mathbb{R}^m$ and we should encounter some "small" $\alpha$ such that $rank(T_xf + \alpha) = m$ for any $x \in \mathbb{R}^n$.
In such a case, the first idea that comes to my mind is to find a (sufficiently) smooth map $G$ such that $$ f+\alpha \text{ is an immersion} \Longleftrightarrow \alpha \text{ is a regular value of }G $$ and then apply Sard's theorem to ensure that there is enough $\alpha$ satisfying our requirement. The problem is that, at the moment, I'm not able to find such a map $G$...
To sum up I look for some hints apply this method or a bit more elaborate ideas (like the use of a transversality theorem). Or maybe more basic ideas, who knows ?
Remark: Where does the hypotesis $n \geq 2m$ play a role ?
This is not true if $f$ is just "differentiable" in the sense of having a derivative everywhere. For instance, let $g:\mathbb{R}\to\mathbb{R}^2$ be a continuous surjection and let $f:\mathbb{R}\to\mathbb{R}^2$ be obtained by integrating $g$, so $f'(t)=g(t)$ for all $t\in\mathbb{R}$. Then no matter what $\alpha:\mathbb{R}\to\mathbb{R}^2$ we add to $f$, there is some $t\in\mathbb{R}$ such that $f'(t)=-\alpha(1)$ and so the derivative of $f+\alpha$ at $t$ will be $0$.
It is true if you assume $f$ is at least $C^2$. Here is one way to prove it.
Let $e_1,\dots,e_m$ be the standard basis for $\mathbb{R}^m$. The idea is then to choose the values of $\alpha(e_i)$ one by one so that $T_x(f)+\alpha$ sends the $e_i$ to linearly independent vectors for all $x$. For instance, to choose $\alpha(e_1)$, you just need $\alpha(e_1)$ to not be equal to $-T_x(f)(e_1)$ for any $x$. You can do this because the map $x\mapsto -T_x(f)(e_1)$ is a $C^1$ map $\mathbb{R}^m\to\mathbb{R}^n$ which cannot be surjective since $m<n$. Moreover, you can choose $\alpha(e_1)$ to be arbitrarily small since the image of this map $\mathbb{R}^m\to\mathbb{R}^n$ has measure $0$ and in particular has empty interior.
The rest of the process is similar: having chosen $\alpha(e_1),\dots,\alpha(e_{i-1})$, you need to choose $\alpha(e_i)$ so that $T_x(f)(e_i)+\alpha(e_i)$ is linearly independent from the $T_x(f)(e_j)+\alpha(e_j)$ for $j<i$, for all $x$. In other words, you need there to be no scalars $c_j$ and no $x$ such that $$\alpha(e_i)=-T_x(f)(e_i)+\sum_{j=1}^{i-1}c_j(T_x(f)(e_j)+\alpha(e_j)).$$ The right side of this equation is a $C^1$ function of $(x,c_1,\dots,c_{i-1})$, going from $\mathbb{R}^{m+i-1}\to\mathbb{R}^n$. Since $i\leq m$, $m+i-1<2m\leq n$ so the image of this map has measure zero and we can pick an arbitrarily small value for $\alpha(e_i)$ that is not in its image.