It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,2),(2,1),(3,1)\}=f$
It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,2),(2,1),(3,1)\}=f$
since $f(2)=f(3)=1, f$ is not one-one, so that $f$ is not invertible.
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