Composition of Mappings, Surjective and Injective

In summary, the composition of mappings is a mathematical operation that combines two functions to create a new function. It is more general than function composition and applies to functions with multiple inputs and outputs. A mapping is surjective if every element in the range is paired with at least one element in the domain, and this is related to the composition of mappings. An injective mapping is a function where each element in the range is paired with at most one element in the domain.
  • #1
gbean
43
0

Homework Statement


a) Let g: A => B, and f: B => C. Prove that f is one-to-one if f o g is one-to-one.
b) Let g: A => B, and f: B => C. Prove that f is onto if f o g is onto.

Homework Equations


a) Since f o g is onto, then (f o g)(a) = (f o g)(b) => a = b.
b) Since f o g is onto, every element in C is an image under f o g.
(f o g)(a) = f(g(a)) = f(b) = c

The Attempt at a Solution



a) Proof by contradiction. Assume that g is not one-to-one.
(f o g)(a) = (f o g)(b)
f(g(a)) = f(g(b))
I don't know where to go from here.
I'm just wondering if this explanation is rigorous enough.
b) Proof by contradiction: Assume that f is not onto.

(f o g)(a) = f(g(a)) = f(b) = c.
However, f is not onto, thus there exists a c in C that is not mapped to by f. Thus by definition, f o g is not onto if not every c in C is an image under f o g. This is a contradiction of the given, hence f must be onto.
 
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  • #2
gbean said:

Homework Statement


a) Let g: A => B, and f: B => C. Prove that f is one-to-one if f o g is one-to-one.
b) Let g: A => B, and f: B => C. Prove that f is onto if f o g is onto.

Homework Equations


a) Since f o g is onto, then (f o g)(a) = (f o g)(b) => a = b.
b) Since f o g is onto, every element in C is an image under f o g.
(f o g)(a) = f(g(a)) = f(b) = c

The Attempt at a Solution



a) Proof by contradiction. Assume that g is not one-to-one.
(f o g)(a) = (f o g)(b)
f(g(a)) = f(g(b))
I don't know where to go from here.



I'm just wondering if this explanation is rigorous enough.
b) Proof by contradiction: Assume that f is not onto.

(f o g)(a) = f(g(a)) = f(b) = c.
However, f is not onto, thus there exists a c in C that is not mapped to by f. Thus by definition, f o g is not onto if not every c in C is an image under f o g. This is a contradiction of the given, hence f must be onto.


For part a), I think you meant that you are going to assume that f is not 1-1.

Anyway, what exactly does this mean for a function to NOT be 1-1? You had it partially correct (though you seem to have been working in the "opposite" direction). It means there is an a =! b such that f(a)=f(b).

So, if f sends a and b to the same element of B, where does g send f(a) and where does g send f(b)? Can you find a contradiction there?
 
  • #3
Wait, I got confused, in a) are you showing that f or g is 1-1?
 
  • #4
Whoops, I wrote down part a wrong. It should be: a) Let g: A => B, and f: B => C. Prove that g is one-to-one if f o g is one-to-one.
 
  • #5
a) Proof by contradiction. Assume that g is not one-to-one.
(f o g)(a) = (f o g)(a')
f(g(a)) = f(g(a'))

If g is not 1-to-1 then there exist distinct a and a' such that g(a) = g(a') = b.

Then f(b) = f(b) => c = c. But this means that (f o g)(a) = c and (f o g)(a') = c, or that there exist distinct a and a' such that (f o g)(a) = (f o g)(a') = c. This is a contradiction as f o g is not 1-to-1, hence g must be 1-to-1.
 

1. What is the composition of mappings?

The composition of mappings is a mathematical operation that combines two functions to create a new function. It is denoted by g∘f, where g and f are the two functions being combined. The result of the composition is a new function that first applies f to the input and then applies g to the result of f.

2. How is the composition of mappings different from function composition?

The composition of mappings is similar to function composition, but it is more general. Function composition only applies to functions with a single input and a single output, while the composition of mappings can be applied to functions with multiple inputs and outputs. In addition, function composition is defined for all possible combinations of functions, while the composition of mappings may not be defined for certain combinations.

3. What does it mean for a mapping to be surjective?

A mapping is surjective if every element in the range of the function is paired with at least one element in the domain. In other words, every output of the function has at least one input that produces it. This is also known as being "onto" because every element in the range is mapped onto by an element in the domain.

4. How is surjectivity related to the composition of mappings?

If both f and g are surjective mappings, then the composition g∘f is also surjective. This is because the range of f is the entire domain of g, and therefore every output of g has at least one input from f that produces it.

5. What is an injective mapping?

An injective mapping is a function where each element in the range is paired with at most one element in the domain. In other words, every output of the function has at most one input that produces it. This is also known as being "one-to-one" because each element in the range is mapped to by only one element in the domain.

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