Is $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ never injective? I have an exercise to prove that $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is not an injective function, and, $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, with n > m is also never an injective function. I made the proof with the 
Rank–nullity theorem, but I think there is a more intuitive way to see this fact
 A: Given no other restrictions on your function $f$, the exercise is incorrect as stated. $\mathbb{R}^n$ and $\mathbb{R}^m$ both have cardinality $\mathfrak{c}$, which by definition means there is a bijection between them. Hence there is an injective mapping with the hypotheses that you want.
(Rank-Nullity can only be applied if $f$ is linear, so it fails here.)
As requested in the comments, now consider the case where $f$ must be $C^1$. Then there is no injective map from $\mathbb{R}^2\rightarrow\mathbb{R}$. Suppose $f$ is such a map. Then $f(\mathbb{R}^2)$ must be connnected because it is the image of a continuous function. Since there are infinitely many elements in the domain, the image of $f$ is a nontrivial interval in $\mathbb{R}$. Now remove some point $a$ from $\mathbb{R}^2$. Then $\mathbb{R}^2\setminus a$ is still connected. But $f(\mathbb{R}^2\setminus a)$ cannot be connected since the image of $f$ was originally an interval. So the image of a continuous function of a connected set is unconnected, which is not possible. This proves our claim. 
A: If $f:\Bbb R^2\to\Bbb R$ is continuous then it cannot be injective.
Hint: We can certainly assume $f$ is not constant. Say $p,q\in\Bbb R^2$ and $f(p)<f(q)$. Choose $\alpha$ with $f(p)<\alpha<f(q)$. You can use the Intermediate Value Theorem to show there is more than one point where  $f=\alpha$. (Think about curves from $p$ to $q$...)
