Suppose f:A→B is a given function. Then there is a function g:f(A)→A sich that

g o f=I_{A}& f o g= I_{ f(A), }if and only if f is a one to one function and g=f^{-1}

Proof(→):

Suppose f is 1-1 and g=f^{-1}then for all a elements of A, if f(a)=b then g(b)=a.

So g(f(a))=g(b)=a.

Thus g o f=I_{A, }Similarly, f(g(b))=f(a)=b for any element b of f(A), so f o g=I_{f(A)}.

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