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X and Y are two sets and f: X->Y If {f(c...

X and Y are two sets and `f: X->Y` If `{f(c)=y; c subX, y subY}` and `{f^(-1)(d)=x;d subY,x sub X`, then the true statement is `(a) f(f^(-1)(b))=b` `(b) f^(-1)(f(a))=a` `(c) f(f^(-1)(b))=b, b sub y` `(d) f^(-1)(f(a))=a, a sub x`

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