I've begun a course in "Real Analysis" recently and I have this trivial exercise. Could someone check if my proof is correct?
Proposition: There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective
Proof: Firstly, we prove injective function $f: A \rightarrow B \Longrightarrow g: B \rightarrow A$ is surjective Suppose $\exists f: A \rightarrow B, $ such that $ f$ is injective, i. e., $ \forall x_{1}, x_{@} \in A, x_{1} \neq x_{2} \rightarrow f(x_{1}) \neq f(x_{2})$.
By hypothesis, $\exists g: B \rightarrow A$ such that $g$ is not surjective. Then,there is at least one $ x \in A $ such that $ \forall y \in B, g(y) \neq x $. But, that is not possible, because if $f$ is injective, then all $x \in A$ correspond to some $y \in B$. Contradiction!
Now, we prove surjective function $g: B \rightarrow A \Longrightarrow f: A \rightarrow B$ is injective. Suppose $g: B \rightarrow A $ is surjective, i. e., $\forall y \in B, \exists x \in A$, such that $ g(y) = x$. By hypothesis, $\exists f: A \rightarrow B$ such that f is not injective. Then, there are $x_{1}, x_{2} \in A$ such that for $x_{1} \neq x_{2}$, there are $f(x_{1}) = f(x_{2})$. By the definition of function, that only could happen, if there is $ y \in B $ such that $ y \notin Dom(g) $. Contradiction!
So, There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective. Q.E.D.
Best Answer
There are several problems with your proof.
Firstly, in both directions of your proof, your statements "By hypothesis, ..." seem to assume that all proofs are by contradiction, and that you must refute the negation of the statement. What you are assuming is not a hypothesis.
Secondly, your negation of the statement you wish to prove is incorrect. The negation of "there exists a surjective function" is not "there exists a function that is not surjective," but rather: "all functions are not surjective." The same goes for your other direction.
You may want to look at the $\underline{inverse}$ of these funtions, and in one case, extend its domain.