# Showing Function is Injective/Surjective

• mikemhz
In summary: Would this show that it is injective?In summary, if a function is injective, then it is also surjective.
mikemhz
Show that for any sets $X, Y , Z$, the canonical function:

$\varphi : (X × Y) × Z \rightarrow X × (Y × Z)$

$(\varphi((x, y), z) = (x,(y, z)))$

is a bijection.

Solution. We can do this by showing that $\varphi$ is injective and surjective..

I can do this by showing $\varphi$ has an inverse (isomorphism theorem). But I would like to know how to show that a function involving cartesian products is injective/surjective.

mikemhz said:
I can do this by showing $\varphi$ has an inverse (isomorphism theorem). But I would like to know how to show that a function involving cartesian products is injective/surjective.
This should be pretty straightforward. Why not start by writing out the definition of injective and surjective for this particular map, i.e., what is it you would need to show in each case?

Would this show that it is injective?

$x, a \in X$
$y, b \in Y$
$z, c \in Z$

$\varphi((x,y),z) = \varphi((a,b),c)$

$\varphi : (x,(y,z)) = (a,(b,c))$

$x=a, y=b, z=c$

End of proof.

How about surjective? To be honest, I'm not 100% what I need to prove in each case. I know that injective is one-to-one (each element in Y has at MOST one element from X mapped to it) and surjective is onto (each element in Y has at LEAST one element from X mapped to it), but I do not know how that translates into the notation.

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Sooo..

$X = \left\{1, 2\right\}$
$Y = \left\{A, B\right\}$
$Z = \left\{I, J\right\}$

$X × (Y × Z) =$
$\left\{(1, (A, I)), (1, (A, J)), (1, (B, I)), (1, (B, J)), (2, (A, I)), (2, (A, J)), (2, (B, I)), (2, (B, J))\right\}$

Take any element from $X × (Y × Z)$:
e.g. $(2, (A, I))$
Show that there is an element from $(X × Y) × Z$ which maps using the function:
$\varphi((2, A), I) = (2, (A, I))$

Therefore $\varphi$ is surjective...? :)

Last edited:
mikemhz said:
Would this show that it is injective?

$x, a \in X$
$y, b \in Y$
$z, c \in Z$

$\varphi((x,y),z) = \varphi((a,b),c)$

$\varphi : (x,(y,z)) = (a,(b,c))$

$x=a, y=b, z=c$

End of proof.
Yes, that's pretty much all there is to it. You may want to include "if and only if" between the statements and write some words to explain the conclusion you are drawing, such as:

$\varphi((x,y),z) = \varphi((a,b),c)$

if and only if

$(x,(y,z)) = (a,(b,c))$

if and only if

$x=a, y=b, z=c$

if and only if

$((x,y),z) = ((a,b),c)$

Therefore $\varphi((x,y),z) = \varphi((a,b),c)$ implies $((x,y),z) = ((a,b),c)$. This shows that the function $\varphi$ is injective.

mikemhz said:
Therefore $\varphi$ is surjective...? :)
No, this isn't a general proof, because you used specific examples of $X$, $Y$, and $Z$. To write a general proof, start by taking an arbitrary element of $X \times (Y \times Z)$. An arbitrary element is of the form $(x,(y,z))$, where $x \in X, y \in Y, z \in Z$. Now you need to find an element of $(X \times Y) \times Z$ which is mapped by $\varphi$ to $(x,(y,z))$. That's all there is to it.

## What is the definition of injective function?

An injective function is a type of function in which each element in the domain maps to a unique element in the range. This means that no two distinct elements in the domain can have the same image.

## How can I show that a function is injective?

To show that a function is injective, you can use the horizontal line test. This involves drawing horizontal lines on the graph of the function and checking if each line intersects the graph at most once. If yes, then the function is injective.

## What is the definition of surjective function?

A surjective function is a type of function in which every element in the range has at least one preimage in the domain. This means that every element in the range is mapped to by at least one element in the domain.

## How can I show that a function is surjective?

To show that a function is surjective, you can use the vertical line test. This involves drawing vertical lines on the graph of the function and checking if each line intersects the graph at least once. If yes, then the function is surjective.

## Can a function be both injective and surjective?

Yes, a function can be both injective and surjective. This type of function is called a bijective function. It means that each element in the domain maps to a unique element in the range, and every element in the range has exactly one preimage in the domain.

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