Assuming that A and B are non-empty, if there is an injective function F : A -> B then there must exist a surjective function g : B -> A Either give a counter-example, or a proof.
A question in my proofs review.
From what I understand we must assume each element of A is carried to a unique element of B (i.e. every value of A is associated with a unique value of B when punched into F). 
We now must prove (or disprove by use of counter-example) that there always exists a surjective function G that carries B -> A.  Surjectivity (from what I understand) means that EVERY element of A (in this case) has AT LEAST one associated B value in terms of the function G.
Will anybody be kind enough to correct any misunderstandings that I may have?
Also, I can't really figure out where to begin with this question, help needed!
 A: I think the key on this one is to to realize that, since $F:A \to B$ is injective, we can define an inverse function $F^{-1}$ on $F(A)$, the image of $A$ in $B$.  For every $z \in F(A)$ is the image of some $y \in A$, that is, $z = F(y)$ for $y \in A$, and since $F$ is injective there is exactly one such $y$.  We set $y = F^{-1}(z)$, on $F(A)$, for $z  \in F(A)$; we note that $F^{-1}:F(A) \to A$ is a surjection, since each $y \in A$ satisfies $y = F^{-1}(F(y))$; we define a function $V:B \to A$ by setting $V(z) = F^{-1}(z)$ for $z \in F(A)$, and $V(z) =a$ for any $a \in A$ when $z \in B - F(A)$; since $F^{-1}:F(A) \to A$ is surjective, $V: B \to A$ is a surjection from $B$ to $A$.  QED.
Hope this helps.  Cheerio,
and as always,
*Fiat Lux!!! *
A: Denoting the injection as $F$: then its inverse, $G$, is a surjection from $F(A) \rightarrow A$. 
(Where $F(A) := \{F(a): a \in A\}$ is the image of $A$ after applying $F$ to it.)
Since $A$ is nonempty, we can pick $a \in A$. 
Extend the domain of $G$ from $F(A)$ to $B$ (if necessary) by mapping all elements in $B-F(A)$ to $a$. 
This (possibly extended) function, $g$, is a surjection $B \rightarrow A$. QED.
