How Do You Prove a Mapping is Surjective?

And that's what we did by solving the system of equations and showing that for every $(X,Y,Z)$ in the co-domain, we can find a unique $(x,y,z)$ in the domain such that $L(x,y,z)=(X,Y,Z)$.
  • #1
rputra
35
0
I have a mapping $L: \mathbb R^3 \rightarrow \mathbb R^3$ as defined by $L(x, y, z) = (x+z, y+z, x+y).$ How do you prove that the $L$ is an onto mapping? I know for sure that $\forall x, y, z \in \mathbb R$, then $x+z, y+z, x+y \in \mathbb R$ too. Then I need to prove that $Im (L) = \mathbb R^3$ the co-domain, but I do not know how to proceed officially.

Any help would be very much appreciated. Thank you for your time.
 
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  • #2
Tarrant said:
I have a mapping $L: \mathbb R^3 \rightarrow \mathbb R^3$ as defined by $L(x, y, z) = L(x+z, y+z, x+y).$ How do you prove that the $L$ is an onto mapping? I know for sure that $\forall x, y, z \in \mathbb R$, then $x+z, y+z, x+y \in \mathbb R$ too. Then I need to prove that $Im (L) = \mathbb R^3$ the co-domain, but I do not know how to proceed officially.

Any help would be very much appreciated. Thank you for your time.

Hi Tarrant! Welcome to MHB! ;)

I assume that should be $L(x, y, z) = (x+z, y+z, x+y)$?

A function is onto iff each point in the co-domain has at least 1 original.

Let's pick a point $(X,Y,Z)$ in the co-domain.
Then to find the originals if any, we need to solve:
$$\begin{cases}x+z=X \\ y+z=Y \\ z+y=Z \end{cases}$$

If we can always find an $(x,y,z)$, the function is onto.
If additionally, there is always exactly one such $(x,y,z)$, the function is also $1-1$ and therefore $bijective$.
 
  • #3
I like Serena said:
Hi Tarrant! Welcome to MHB! ;)

I assume that should be $L(x, y, z) = (x+z, y+z, x+y)$?

A function is onto iff each point in the co-domain has at least 1 original.

Let's pick a point $(X,Y,Z)$ in the co-domain.
Then to find the originals if any, we need to solve:
$$\begin{cases}x+z=X \\ y+z=Y \\ z+y=Z \end{cases}$$

If we can always find an $(x,y,z)$, the function is onto.
If additionally, there is always exactly one such $(x,y,z)$, the function is also $1-1$ and therefore $bijective$.
Thanks for responding to my posting and thanks also to your correction. I have made the correction. In hindsight, I think I can also prove it the traditional ways, that is first proving that the range $Im(L) \subset \mathbb R^3$ the co-domain, and then the other way around, proving that the co-domain $\mathbb R^3 \subset Im(L)$ the range. Then these two of them lead to the range $Im(L) = \mathbb R^3$ the codomain. Remember the mapping we have is $L: \mathbb R^3 \rightarrow \mathbb R^3.$

Thank you again.
 
  • #4
Tarrant said:
Thanks for responding to my posting and thanks also to your correction. I have made the correction. In hindsight, I think I can also prove it the traditional ways, that is first proving that the range $Im(L) \subset \mathbb R^3$ the co-domain, and then the other way around, proving that the co-domain $\mathbb R^3 \subset Im(L)$ the range. Then these two of them lead to the range $Im(L) = \mathbb R^3$ the codomain. Remember the mapping we have is $L: \mathbb R^3 \rightarrow \mathbb R^3.$

Thank you again.

That's the same thing.
$\operatorname{Im}(L) \subset \mathbb R^3$ is implicit from the function definition.
So what we need to prove is indeed $\mathbb R^3 \subset \operatorname{Im}(L)$, meaning every point in the co-domain has to have at least one original.
 

FAQ: How Do You Prove a Mapping is Surjective?

What is a surjective mapping?

A surjective mapping, also known as a surjection, is a type of function that has a range that is equal to its codomain. This means that every element in the codomain has at least one corresponding element in the domain.

How do you prove that a mapping is surjective?

To prove that a mapping is surjective, you must show that every element in the codomain has at least one corresponding element in the domain. This can be done by showing that for every element in the codomain, there exists at least one element in the domain that maps to it.

What is the difference between surjective and injective mappings?

A surjective mapping ensures that every element in the codomain has at least one corresponding element in the domain, while an injective mapping ensures that every element in the domain has at most one corresponding element in the codomain.

Can a function be both surjective and injective?

Yes, a function can be both surjective and injective. This type of function is called a bijective function, meaning that every element in the domain has one and only one corresponding element in the codomain.

Why is proving a mapping is surjective important?

Proving that a mapping is surjective is important because it ensures that every element in the codomain is being mapped to by at least one element in the domain. This is necessary for the function to be considered "onto" or "covering" the entire codomain, and is an important aspect of understanding functions and their properties.

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