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If `f:A->B, g:B->C` are bijective functions show that `gof:A->C` is also a bijective function.A. `f^(-1)og^(-1)`B. `fog`C. `g^(-1)of^(-1)`D. `gof` |
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Answer» Correct Answer - A Given that, `f:A to B` and `g :B to C` be the bijective functions. `(gof)^(-1)=f^(-1)og^(-1)` |
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