Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a vector-field where the function $f_i$ for each dimension $i$ is continuous and strictly increasing. Are these assumptions enough to conclude that the vector-field $f$ is surjective and injective?
As I can understand from the comments, the term "strictly increasing" is not well-defined for vector-fields. What I mean is that if $\vec{x}$ and $\vec{y}$ are two vectors, then I say that $\vec{x} < \vec{y}$ if $\exists i: x_i < y_i \wedge \forall j \neq i: x_j \leq y_j$. So by strictly increasing $f_i$ I mean that if $\vec{x} < \vec{y}$ then $f_i(\vec{x}) < f_i(\vec{y})$.
Monotonicity of functions $f:{\mathbb R}^n \to {\mathbb R}^n$ often refers to the order on ${\mathbb R}^n$ defined by the cone $K=[0,\infty)^n$ via $$ x \le y :\iff y-x \in K \quad (\iff x_k\le y_k, k=1,\dots,n). $$ This is an order relation on ${\mathbb R}^n$, but in case $n\ge 2$ its not a total ordering (for example $(1,0)$ and $(0,1)$ cannot be compared). As usual $$ x<y :\iff x \le y \wedge x\not= y, $$ that is $x < y$ means $x_k\le y_k$ for $k=1,\dots,n$ and there is at least one coordinate $l$, say, with $x_l < y_l$. Moreover sometimes $$ x \ll y :\iff x_k<y_k, k=1,\dots,n $$ is used. For example $f: {\mathbb R}^2 \to {\mathbb R}^2$, $f(x_1,x_2) =(x_1+x_2,x_1+x_2)$ ist strongly increasing in the sense that $$ x < y ~ \Rightarrow ~ f(x) \ll f(y), $$ but $f$ is neither surjective (for example $(0,1)$ cannot be reached by $f$) nor injective ($f(0,1)=f(1,0)$). Note that $f(x) \ll f(y)$ implies $f(x) < f(y)$, so $f$ is strictly increasing.