1. ## Injectivity and Surjectivity

Let $\displaystyle f: X \to Y$ be a function. Prove that (i) $\displaystyle f$ is injective $\displaystyle \Leftrightarrow \overrightarrow{f}$ is injective $\displaystyle \Leftrightarrow \overleftarrow{f}$ is surjective.

(ii) $\displaystyle f$ is surjective $\displaystyle \Leftrightarrow$ $\displaystyle \overleftarrow{f}$ is surjective $\displaystyle \Leftrightarrow \overleftarrow{f}$ is injective.

So for (i) $\displaystyle \forall x_1, x_2 \in X, (f(x_1) = f(x_2) \Rightarrow x_1 = x_2 )$. So $\displaystyle \overrightarrow{f}(A) = \{f(x)| x \in A \}$ for $\displaystyle A \in \mathcal{P}(X)$. Say $\displaystyle x_{1} \in A$ and $\displaystyle x_2 \in A$. Then $\displaystyle f(x_1) = f(x_2) \Rightarrow x_1 = x_2$. $\displaystyle \overleftarrow{f}(B) = \{x \in X| f(x) \in B \}$ for $\displaystyle B \in \mathcal{P}(Y)$ and so $\displaystyle \forall y \in Y, \exists x \in X, f(x) = y$ which implies surjectivity.

So then you go in the opposite direction to complete the proof?

Is it a similar case for (ii)?

2. Did I get the general proof correct?