Show that an onto function $f:\{1{,}2{,}3\} \rightarrow\{1{,}2{,}3\}$ is always one-one.

Step 1: Understand the Definitions

We are given a function $f$ from set $A = \{1{,}2{,}3\}$ to set $B = \{1{,}2{,}3\}$. We need to show that if $f$ is onto, it must also be one-one. Let's recall the definitions of onto and one-one functions.

Step 2: Definition of Onto Function

An onto (or surjective) function means that every element in the codomain (set $B$) has at least one corresponding element in the domain (set $A$) that maps to it. In simpler terms, the range of the function is equal to its codomain.

Step 3: Apply Onto Condition to the Given Sets

Since $f$ is an onto function from $A=\{1{,}2{,}3\}$ to $B=\{1{,}2{,}3\}$, every element in $B$ must be an image of some element in $A$. This means that the set of images $\{f(1), f(2), f(3)\}$ must be exactly $\{1{,}2{,}3\}$.

Step 4: Definition of One-One Function

A one-one (or injective) function means that distinct elements in the domain map to distinct elements in the codomain. In other words, if two elements in the domain have the same image, then they must be the same element.

Step 5: Prove One-One using Onto Property

We know that the range of $f$ is $\{1{,}2{,}3\}$. This means that $f(1)$, $f(2)$, and $f(3)$ must take on the values $1$, $2$, and $3$ in some order. For this to happen, $f(1)$, $f(2)$, and $f(3)$ must all be distinct values. If any two of them were the same (e.g., $f(1)=f(2)$), then the range would contain only two distinct values, which contradicts the fact that the range is $\{1{,}2{,}3\}$. Since $f(1)$, $f(2)$, and $f(3)$ are distinct, it implies that distinct elements in the domain map to distinct elements in the codomain, thus $f$ is one-one.

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