Show that a one-one function $f:\{1{,}2{,}3\} \rightarrow\{1{,}2{,}3\}$ must be onto.

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 one-one, it must also be onto.

Step 2: Analyze One-One Property

A function is one-one (or injective) if distinct elements in the domain map to distinct elements in the codomain. This means that if $f(1)$, $f(2)$, and $f(3)$ are the images of the elements in the domain, they must all be different.

Step 3: Determine the Images

Since $f$ is one-one, the images $f(1)$, $f(2)$, and $f(3)$ must be three distinct elements. The codomain $B$ also has exactly three elements: $1, 2, 3$.

Step 4: Relate Images to Codomain

Since $f(1)$, $f(2)$, $f(3)$ are three distinct elements and they must belong to the codomain $B = \{1{,}2{,}3\}$, the set of images \{f(1), f(2), f(3)\} must be exactly equal to the codomain \{1,2,3\}.

Step 5: Conclude Onto Property

A function is onto (or surjective) if every element in the codomain has at least one corresponding element in the domain. Since the set of images (the range) is equal to the codomain, every element in the codomain $B$ is an image of some element in the domain $A$. Therefore, the function $f$ is onto.

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