Injectivity and Surjectivity

**shilz222** · August 23rd 2007, 08:41 PM

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

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

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

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

Is it a similar case for (ii)?

**shilz222** · August 24th 2007, 11:25 AM

Did I get the general proof correct?

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