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.

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- February 12th 2010, 07:32 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*. - February 12th 2010, 09:34 AMDrexel28
This is unreadable.

It should read for the first one

Quote:

for every

__Proof:__

:

(this is always true). So,

**Lemma:**

**Proof**: Suppose that and let . Since there exists some such that . But, since we see that and so which condtradicts 's injectivity. The conclusion follows

Now, then

so

from where it follows that

Let then and so from where it follows that - February 12th 2010, 11:18 AMrezonathank
thank you