Hey guys thanks for the help, I think I figured out the answer thanks to you guys.
If I'm understanding correctly the power set of {abc} will map to an integer based on the cardinality of each of the subsets thus
{null set} maps to 0, {a} maps to 1, {b} maps to 1, (a,b} maps to 2, {a,b,c} maps to 3 and so forth.
Thus we have more than one x value mapping the same y value,{a} and {b} both map to 1 so it is not a one to one function.
Furthermore, if this function is to map to the set of all integers, all of the integers are not covered as we only use 0,1,2 and 3. Therefore it is not onto either.
Hope that helps anyone who needs assistance on this question in the future.
Let me know if I made any mistakes.
Best Answer
Induct on $|X|$. The base case $|X|=1$ is obvious since then there is only one function $X\rightarrow X$.
Now, suppose inductively that $|E|\leq n$ implies that every injective $E\rightarrow E$ is surjective. Suppose $|X|=n+1$ and let $f:X\rightarrow X$ be injective. Seeking a contradiction, suppose $f$ is not surjective so $|f(X)|\leq n$. Then $g:f(X)\rightarrow f(X)$ given by $g(t)=f(t)$ is injective and the inductive hypothesis implies $g$ is surjective. That is, $g(f(X))=f(X)$ so for every $y\in X$ there exists an $x\in X$ such that $f(f(x))=f(y)\Rightarrow f(x)=y$. Thus $f$ is surjective, a contradiction. Hence $f$ is surjective and this closes the induction.