Injective function for $F^n\rightarrow F^m$ $X$ and $Y$ are $F$-linear and $\dim_FX=n$ and $\dim_FY=m$.
If $n\leq m$, does there exist an $F$-linear injective function $f:X\rightarrow Y$?
Well of course, but how do i notate that?
$$f\begin{pmatrix}x_1\\x_2\\ \vdots\\x_n\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\ \vdots\\x_n\\0\\\vdots\\0\end{pmatrix}$$
And am I right that this is from $F^n\to F^m$?
Note: for me it is not clear wether $x_i$ is element of Y or not, should i instead map $x_i \to y_i$?
Also, does $n\leq m$ imply that at least $n$ is finite?
 A: Hint: 
Let $\{x_i\}_{i=1}^n$ be a basis of $X$. Let $\{y_i\}_i$ be a basis of $Y$. 
Prove that the unique linear transformation that sends $x_i\mapsto y_i$ is an injection (further hint, use that $\{y_i\}$ is a basis of $Y$). 
Regarding you questions: Yes, $x_i$ may not be an element of $Y$, so it makes little sense to use it as such.
Also, we could have $n\leq m$ even when $n$ is not finite (we could have $n=\aleph_0,\, m=\mathfrak c$).

The transformation is indeed $T\,x_i=y_i$. It's a basic theorem in linear algebra that a linear transformation is completely determined by its action on a basis (that's why we can talk about the linear transformation here).
Let $\{y_i\}$ be as defined above. Let $v_1,v_2\in X$ and suppose $T \,v_1=T\, v_2$. We know that $v_1=\sum c_i x_i$ and $v_2=\sum k_i x_i$ for some $c_i,k_i$. 
From linearity, it follows that $T \, v_1 = \sum c_i y_i$ and $T\, v_2=\sum k_i y_i$. 
Here we use that $\{y_i\}$ is a basis, this gives us that the representation of any vector $u$ as a sum $\sum r_i y_i$ has a unique combination of $r_i$'s, so we get that $c_i=k_i$ and finally, $v_1=v_2$, thus $T$ is injective.
A: Well, according to the dimensions, assume that $A=\{v_1,\cdots, v_n\}$ and $B=\{u_1,\cdots, u_m\}$ are the basis of $X$ and $Y$, respectively. 
Now, for each $x\in X$, we know that there are $c_1, c_2, \cdots, c_n$ such that $x=\sum_{i=1}^{n}c_iv_i$. So, it is enough to define $f(x)=\sum_{i=1}^{n}c_iu_i$. 
It is simple to show that it is an injection.
This way, you can be sure that the image of $f$ is subset of $Y$.
