If we have $f$ is one-to-one, why can we conclude that $n\mathbb{Z}_m=\mathbb{Z}_m? $ Suppose $m,n \in \mathbb{Z},m,n\geq1.$ Define a map $$f:\mathbb{Z}_m \rightarrow \mathbb{Z}_m$$ where $[x] \rightarrow [nx]$ If we have $f$ is one-to-one, why can we conclude that $n\mathbb{Z}_m=\mathbb{Z}_m? $

Any one to one function $f$ from a set $A$ with $m$ elements to a set $B$ with $m$ elements is onto. And any onto function is one to one. 
This does not hold for infinite sets of the same cardinality. 
The proof for finite sets is a matter of counting. Since $f$ is one to one, the values of $f$ at the $m$ elements of $A$ are all different. So $f$ takes on $m$ distinct values. Since $B$ only has $m$ elements, the values of $f$ must include all the elements of $B$.
Think of it this way. Let $A$ be a set of $m$ women, and let $B$ be a set of $m$ men. Each woman $a$ chooses a man $f(a)$, with the rule that no two women can choose the same man. (That says $f$ is one to one.) Will there be a man who remains unchosen? Certainly not. Thus the function $f$ is onto.  
