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If function f is an isomorphism from semigroup (S, *) to (T, *'), then prove that $f^{-1}$ ia an isomorphic from (T, *') to (S, *).

written 7.9 years ago by teamques10 ★ 68k

modified 2.9 years ago by sagarkolekar ★ 11k

discrete mathematics

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written 7.9 years ago by teamques10 ★ 68k

Let a' and b' $\in$ T

Since $f$ is onto, we can find elements $a$ and $b$ in $S$ such that $$(\because Ran(f)=(s,*))$$

$f(a)=a'$ and $f(b)=b'$

Then, $a=f^{-1}(a')$ and $b=f^{-1}(b')$

Now, $f^{-1}(a'*'b')$ $=f^{-1}(f(a)*'f(b')) \\ =f^{-1}(f(a*b)) \\ =(f^{-1} \circ f)(a*b) \\=a*b=f^{-1}(a')*f^{-1}(b')$

Hence, $f^{-1}$ is an isomorphism

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