For every functionf, subsetAof the domain and subsetBof the codomain prove that Iffis injective we haveA=f−1(fA) and iffis surjective we havef(f−1B) =B.

Printable View

- Feb 12th 2010, 06:32 AMrezonainjective 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*−1*B*) =*B*. - Feb 12th 2010, 08:34 AMDrexel28
This is unreadable.

It should read for the first one

Quote:

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

__Proof:__

$\displaystyle \implies$:

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

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

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

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

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

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

$\displaystyle \Leftarrow:$ Let $\displaystyle f(x)=f(y)$ then $\displaystyle f(\{x\})=\{f(x)\}=\{f(y)\}=f(\{y\})$ and so $\displaystyle \{x\}=f^{1}f(\{x\})=f^{-1}f(\{y\})=\{y\}$ from where it follows that $\displaystyle x=y$ - Feb 12th 2010, 10:18 AMrezonathank
thank you