For the first four problems, let f: X rightarrow Y and g: Y rightarrow Z. If f i
ID: 3120105 • Letter: F
Question
For the first four problems, let f: X rightarrow Y and g: Y rightarrow Z. If f is not injective, is it possible that g o f is injective? If so, give an example (perhaps with an arrow diagram); if not, prove why it is impossible (and it might help to write down that statement formally). If g is not injective, is it possible that g o f is injective If so, give an example (perhaps with an arrow diagram); if not, prove why it is impossible (and it might help to write down that statement formally) The text defines f^-1 for any function, then goes on to say that f^-1 is only a function when f is a bijection. Why is that? Specifically: if f is not injective, then what part of the definition of function does f^-1 violate? If f is not surjective, then what part of the definition of function does f^-1 violate?Explanation / Answer
1) if f is injective, there exists X1 x2 such that f(x1) = f(x2);
gof(X1) = g(f(x1))=g(f(x2)) as f(x1) = f(x2)
So gof is not injective if f is not injective. I
2) if g is not injective then also it's possible that gof is injective. if f is injective but not surjective.
So if f does not maps to those elements in g for which g is injective. Then gof can be injective.
Suppose x1 x2 are two only element in Y for which g(x1) = g(x2);
If for any value x in f(x) it is not equal to x1 or x2, then gof will be an injective function.
3) for f-1 to exist f o f-1
And f-1 o f must exist. And equal to x:
if f:X-->Y then f inverse maps from Y-->X
For f o f inverse and f inverse o f to exist simultaneously f must be surjective.
If f is not surjective then f inverse will not exists as a function, because there will be elements in domain of f inverse
That will not map to any elements in codomain of f inverse which violates def of function.