Question Statement

Given that $A=\{0,1,2,3\}$, $B=\{p,q,r,s\}$ and the function $f$ is defined by the set of ordered pairs: $f=\{(0,p),(1,q),(2,r),(3,s)\}$

Check whether the function is one-to-one, onto and/or into.

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Background and Explanation

A function is one-to-one (injective) if every distinct element in the domain maps to a distinct element in the codomain. A function is onto (surjective) if the range of the function is equal to its codomain, meaning every element in the codomain is mapped to by at least one element in the domain.

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Solution

To determine the nature of the function, we first list the mappings provided by the set of ordered pairs:

• $f(0)=p$
• $f(1)=q$
• $f(2)=r$
• $f(3)=s$

One-to-One (Injective) Check

A function is one-to-one if for every $x_1,x_2\in A$, $f(x_1)=f(x_2)$ implies $x_1=x_2$. In this case, we observe that:

• $0,1,2,3$ are all distinct elements of $A$.
• $p,q,r,s$ are all distinct elements of $B$.

Since each value in the domain $A$ maps to a different, unique value in the codomain $B$, the function $f$ is one-to-one.

Onto (Surjective) Check

A function is onto if the Range of the function is equal to the Codomain $B$.

• The Codomain is given as $B=\{p,q,r,s\}$.
• The Range (the set of actual outputs) is $\{f(0),f(1),f(2),f(3)\}=\{p,q,r,s\}$.

Since $\text{Range}(f)=B$, every element in $B$ has a pre-image in $A$. Therefore, the function $f$ is onto.

Into Check

A function is called an "into" function if there is at least one element in the codomain that does not have a pre-image in the domain (i.e., the Range is a proper subset of the Codomain). Because this function is onto ($\text{Range}=\text{Codomain}$), it is not an into function.

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Key Formulas or Methods Used

• One-to-one (Injective): $f(x_1)=f(x_2)⟹x_1=x_2$
• Onto (Surjective): $\text{Range}(f)=\text{Codomain}$
• Into: $\text{Range}(f)\subset \text{Codomain}$ (Range is a proper subset)

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Summary of Steps

1. List the image of each element of the domain $A$ based on the given ordered pairs.
2. Verify if any two different inputs result in the same output; if not, the function is one-to-one.
3. Determine the Range of the function by collecting all the output values.
4. Compare the Range to the Codomain $B$. If they are identical, the function is onto.
5. Conclude that since it is onto, it cannot be strictly "into."