MHB How Do You Prove a Mapping is Surjective?

[rputra]

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.

 

[I like Serena]

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 $bijective$.

 

[rputra]

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.

 

[I like Serena]

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.