If every fuction $f:A \mapsto B$ with $A\subseteq B, f(A)=B$ is not injective(1-1) then $A\supset B$

Question

How can we prove this statement:

If for every fuction $f:A \mapsto B$ with $A\subseteq B, f(A)=B$ is not injective(1-1) then $A\supset B$

Answer 1

I believe you've either mis-stated the result or that the result given to you was mis-stated. Instead, it seems that it should be stated in one of the two following forms:

1. If $A\supseteq B,$ and for every surjection $f:A\to B$ we have that $f$ is non-injective, then $A\supset B.$
2. If $A\subseteq B,$ and for every surjection $f:A\to B$ we have that $f$ is non-injective, then $A\subset B.$

The upshot, here, is that the hypotheses above state that there is no bijection $A\to B.$ Consequently, we can't have $A=B$ (for if that were true, then the identity function would contradict the hypotheses), so $A\supseteq B$ would imply that $A\supset B,$ and $A\subseteq B$ would imply that $A\subset B.$

---

Noting that it is impossible for $A\subseteq B$ and $A\supset B$ to be simultaneously true, then in order for the implication to be true, we would have to prove that the hypotheses are necessarily false. However, this will not be possible. Consider $A=\{0\}$ and $B=\{0,1\}.$ Note that $A\subseteq B,$ and that every surjection $f:A\to B$ is an injection, because there aren't any surjections $f:A\to B$. However, it certainly isn't true that $A\supset B.$