# injective and surjective

• Feb 12th 2010, 07:32 AM
rezona
injective and surjective
For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.
• Feb 12th 2010, 09:34 AM
Drexel28
Quote:

Originally Posted by rezona
For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.

It should read for the first one
Quote:

$f\text{ is injective }\Longleftrightarrow f^{-1}f(A)=A$ for every $A\in\text{Codom }f$
Proof:

$\implies$:

$x\in A\implies f(x)\in f(A)\implies x\in f^{-1}ff(A)$ (this is always true). So, $A\subseteq f^{-1}f(A)$

Lemma: $f(x)\in f(A)\implies x\in A$

Proof: Suppose that $x\notin A$ and let $y=f(x)$. Since $y\in f(A)$ there exists some $a\in A$ such that $f(a)=y$. But, since $x\notin A$ we see that $x\ne a$ and so $f(x)=y=f(a)$ which condtradicts $f$'s injectivity. The conclusion follows $\blacksquare$

Now, then $x\in f^{-1}f(A)\implies f(x)\in f(A)\implies x\in A$

so $f^{-1}f(A)\subseteq A$

from where it follows that $A=ff^{-1}(A)$

$\Leftarrow:$ Let $f(x)=f(y)$ then $f(\{x\})=\{f(x)\}=\{f(y)\}=f(\{y\})$ and so $\{x\}=f^{1}f(\{x\})=f^{-1}f(\{y\})=\{y\}$ from where it follows that $x=y$
• Feb 12th 2010, 11:18 AM
rezona
thank
thank you