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# Mappings watch

1. Let f:A -> B and g:A->B

Prove that
i) g o f injective => f injective
ii) g o f surjective => g surjective

Either prove or disprove by counterexample the following statement:

g o f = I_A => f o g = I_B

Thanks!
2. (Original post by CharlyH)
Let f:A -> B and g:A->B

Prove that
i) g o f injective => f injective
ii) g o f surjective => g surjective
(i)
Let f:A -> B and g:B -> C? In g o f we have g acting on the elements in the image of f, and this won't necessarily be defined as you've written it unless B is a subset of A.

In any case, we can easily prove the statement (and not worry about the sets, provided we assume that the functions are well defined on what they are acting on).

Recall that if a function is injective then distinct elements are mapped to distinct images. i.e.
If f isn't injective s.t. and . Hence since g is a map, . But since , f o g hence isn't injective. Thus we have shown that if f isn't injective then g o f isn't injective and so can conclude g o f injective => f injective.

(ii)
This can be done in a similar manner.
3. (Original post by Gaz031)
(i)
Let f:A -> B and g:B -> C? In g o f we have g acting on the elements in the image of f, and this won't necessarily be defined as you've written it unless B is a subset of A.

In any case, we can easily prove the statement (and not worry about the sets, provided we assume that the functions are well defined on what they are acting on).

Recall that if a function is injective then distinct elements are mapped to distinct images. i.e.
If f isn't injective s.t. and . Hence since g is a map, . But since , f o g hence isn't injective. Thus we have shown that if f isn't injective then g o f isn't injective and so can conclude g o f injective => f injective.

(ii)
This can be done in a similar manner.

I'm not familiar with your notation. Is I_A the identity map from A to A and I_B the identity map from B to B? The fact there there are left and right inverses should make you suspect that the statement is incorrect.
sorry yeah i made a mistake in the typing, it is supposed to be g: B->A.

yes, also the I_A is the identity mapping from A to A
4. Just for me to try to explain things a bit more, recall that the contrapositive is a method of proof in which when we want to prove we instead prove and since and are equivelant we deduce that is true.

To prove that 'g o f surjection => g surjection' on the two functions you've defined, you can use the contrapositive again. It might be useful for you to recall the definition of a surjection in the more formal form:
We say is a surjection if s.t. .
Then you can say that if isn't a surjection then s.t. s.t. .
From there you can show f o g isn't surjective and hence establish the desired proof via contrapositive.

For the final part of the question it's probably easier to write down a map such as:
f: (a1, a2, a3 -> b1, b2, b3) defined by f(a1)=b1, f(a2)=b2, f(a3)=b3
g: (b1, b2, b3, b4 -> a1, a2, a3) defined similar to above, but with b4 going to some element distinct from a1, a2, a3.
Then clearly g o f is the identity for the set {a1, a2, a3}
However, f o g would map b4 to b3.
5. As your example, f: A -> B, A = {a1, a2, a3}, B = {b1, b2, b3}
Then for the next one g : B -> A, B = {b1, b2, b3, b4}, A = {a1, a2, a3}
Should A and B be two distinct sets?
6. (Original post by BCHL85)
As your example, f: A -> B, A = {a1, a2, a3}, B = {b1, b2, b3}
Then for the next one g : B -> A, B = {b1, b2, b3, b4}, A = {a1, a2, a3}
Should A and B be two distinct sets?
While A and B don't have to be distinct sets, there is a problem in that I've rather stupidly changed the set B half way through the question, forgetting the condition that B maps back to A.
Hence the 'counter example' isn't quite correct.
7. yeah.. I meant to ask A, B in your two examples should be distinct or not
8. If g o f = I_A then f must be injective
And if f o g = I_B then f must be surjective?
So can't have g o f = I_A => f o g = I_B
Is it right? Ofcourse I must think of the counter example .. but haven't found out

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Updated: November 13, 2005
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