Practicing Success
Practicing Success
CUET Preparation Today
| "Let f : A →B be a function. Assume there is a function g : B → A satisfying fog(x) = x then" |
f is invertible
f is neither one one nor onto
f is onto but need not be one one
Nothing can be decided about f
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| f is onto but need not be one one |
| For showing f is onto , start with y in B and consider g(y). For one one , try to cook up some examples by taking f and g finite sets.` |
