00:01
Hello everyone, in this question we are given that suppose that f is a bijective and f composition g is in the first part we have to prove that g is an injection if and only if f composition g is.
00:16
Now let us first define injection.
00:22
So let f from a to b is a function then we can say f is an injection if and only if whenever f of a is equal to f of b so that we will have a is equal to b.
00:51
Now using this definition let us prove the first part.
00:55
So according to the given condition we have f is bijection that is f is injective and surjective.
01:12
So it satisfies both the condition.
01:19
Now let us take g is an injection.
01:28
So consider f of g of a is equal to f of g of b so that this can be written as f of g of a is equal to f of g of b which implies that g of a is equal to g of b since we have already said that f is injection and from this we can say a is equal to b since g is an injection.
02:06
So from this we can say that f of g of a is equal to f of g of b implies a is equal to b which implies f of g is injection.
02:27
Now let us prove the converse part.
02:33
So conversely let us consider g of a is equal to g of b where f of g of a is equal to f of g of b since it is a one -one function.
02:54
Now from this we can write f composition g of a is equal to f composition g of b which implies a is equal to b since f composition g is an injection.
03:09
So from this we can say that g of a is equal to g of b implies a is equal to b since g is an injection.
03:22
So hence we prove that g is injection if and only if f of g is.
03:27
Now let us prove the second part...