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I don't understand it. See for it to be isomorphism you need to it to be homomorphism between them. I can see how this is trivial from what's worked out. I can sort of agree that surjection condition is obvious. However, don't see how he proves injection.

If $af=bf$, then I see that $a-b \in Kerf$. However, I don't see how this gives the injection condition. Looking at the proofwiki doesn't help as it uses a totally different proof. Plus can't get algebra 2 notes that I used last year.

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If $a - b$ is in the kernel, then $[a] = [b]$. This establishes well-definition and injectivity. –  user18063 Dec 28 '11 at 16:33

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up vote 4 down vote accepted

If $a-b\in \ker f$ then $a,b$ define the same cosets, which means you started with the same elements, which is what injectivity tells you should happen!

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Oh, I feel sort of stupid now. Thanks. –  simplicity Dec 28 '11 at 16:41
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You gotta feel stupid at least once in a while, right? Don't worry about that. =) We all do anyway. –  Patrick Da Silva Dec 28 '11 at 16:57

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