# proof a function is an isomorphism

When we prove a function is an isomorphism, we need to prove it's a bijection and it's closed under an operation. In one example I had no problem proving the first part, but in the second part, I proved that $f^{-1}(ab)=f^{-1}(a)f^{-1}(b)$, so my question is does it also follow that $f(ab) = f(a)f(b)$?

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Using what you have already shown:$$f^{-1}(f(a)f(b)) = f^{-1}(f(a)) f^{-1}(f(b)) = ab \\ f(a)f(b) = f(f^{-1}(f(a)f(b))) = f(ab)$$

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Yes write $a=f(a')$ and $b=f(b')$ then $$f^{-1}(ab)=f^{-1}(a)f^{-1}(b)\iff f^{-1}(f(a')f(b'))=a'b'\iff f(a')f(b')=f(a'b')$$

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Yes, a bijective homomorphism's inverse is also a homomorphism.

Also, it does not make sense to say that a function is closed under an operation. A function is compatible with an operation. A subset is closed under an operation.

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If it is true that $f^{-1}(ab)=f^{-1}(a)f^{-1}(b)$, then we have $$f(ab)=f^{-1}(f^{-1}(ab))=f^{-1}(f^{-1}(a)f^{-1}(b))=f^{-1}(f^{-1}(a))f^{-1}(f^{-1}(b))=f(a)f(b)$$

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if $f:E\to F$, $f^{-1}\circ f^{-1}$ is not defined – mookid Apr 13 '14 at 23:19