Prove rk$B$ $\le$ rk$A$ where A and B are free, abelian and finitely generated groups. Let $A$ and $B$ be free abelian, finitely generated groups. Let $f:A \to B$ be an epimorphism. Prove rk$B$ $\le$ rk$A$. I could really use a verification. That is a question from my exam today. 
$proof$: Let rk$A=n$ and rk$B=m$. Suppose $m>n$. Let $b_1,...,b_m$ be a basis of $B$. $f$ is surjective, hence, there exist $x_1,...,x_m\in A$ such that $f(x_i)=b_i\space \forall 1\le i\le m$. Therefore $f(x_1),...,f(x_m)$ are linearly independent. Suppose $x_1,...,x_m$ are linearly dependent. Then there exists $x_j$ such that $x_j=a_1x_{i_1}+...+a_{m-1}x_{i_{m-1}}$ where $x_{i_{k}}\in\{x_1,x_2,...,x_m\}\setminus\{x_j\}$. But that means $f(x_j)=f(a_1x_{i_1}+...+a_{m-1}x_{i_{m-1}})=f(x_j)=a_1f(x_{i_1})_+...+a_{m-1}f(x_{i_{m-1}})$ $\Rightarrow$ $f(x_1),...,f(x_m)$ are linearly dependent. A contradiction. Therefore, $x_1,...,x_m$ are linearly independent. Let us look at $<x_1,...,x_m>$. It is a group of rank $m$. On the other hand, it is a subgroup of $A$ and therefore $m\le n$. A contradiction. Therefore, $m\le n$. 
 A: Linear independence is usually presented in the context of a vector space. Here, the issue is that the $a_i$ are not necessarily invertible. For example, consider $x_1 = 4$ and $x_2 = 6$ in $\mathbb{Z}$. Although $3x_1 - 2x_2 = 0$, it's not true that $x_1 = ax_2$ or $x_2 = a' x_2$ for any $a, a'\in \mathbb{Z}$.
You have the right idea, though. Consider $A\otimes \mathbb{Q}, B\otimes \mathbb{Q}$ to make this argument work, or use the fact that $\operatorname{rk}(A) = \dim_{\mathbb{Q}} (A\otimes \mathbb{Q})$. If you're not familiar with tensor products, then note that for any generating set $x_1, \dots, x_n$ of $A$, the $n$ elements $f(x_1), \dots, f(x_n)$ generate $B$; you probably have some result allowing you to conclude that $\operatorname{rk}(B) \leq n$. 
Specifically, if you're using the definition that $\operatorname{rk}(X)$ is the size of a maximal linearly-independent subset of $X$ (for $X$ a f.g. abelian group), then proceed as follows. If $x_1, \dots, x_n$ is a linearly-dependent subset of $A$, then $f(x_1), \dots, f(x_n)$ is such a subset of $B$, for $\sum \alpha_i x_i = 0$ implies that $\sum \alpha_i f(x_i) = f(\sum \alpha_i x_i) = 0$. Thus if $f(x_1), \dots, f(x_n)$ are linearly independent, so are $x_1, \dots, x_n$. The result follows from the fact that $f:A \to B$ is surjective.
A: You're correct, but you're overdoing.
If $\{y_1,y_2,\dots,y_m\}$ is a  linearly independent set in $B$, then, taking $y_i=f(x_i)$ (for $i=1,2,\dots,m$), we have that $\{x_1,x_2,\dots,x_m\}$ is a linearly independent set in $A$, because a linear dependence relation between $x_1,\dots,x_m$ would give one between $y_1,\dots,y_m$.
Since the rank of $A$ is the number of elements in a maximal linearly independent set, you have proved that $\operatorname{rk}A\ge\operatorname{rk}B$.
Note that this doesn't use the fact that $A$ and $B$ are free and, indeed, the theorem is true for all (finitely generated) abelian groups $A$ and $B$ with an epimorphism $f\colon A\to B$.
In the case when $A$ and $B$ are free, you can say much more. Since $B$ is free, you can take $\{y_1,\dots,y_m\}$ to be a basis for $B$, so $\langle x_1,\dots,x_m\rangle$ splits in $A$ and, if $C$ is a complement, we know that $C$ is free, so $\{x_1,\dots,x_m\}$ can be completed to a basis of $A$ by adding to it a basis of $C$.
