Onto Functions

1 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

Idea

An onto, or surjective, function reaches every element of its codomain. The proof starts with an arbitrary target and constructs an input that reaches it.

Visual guide

VISUAL GUIDEEvery codomain value is reached

Onto, with a repeated output

All elements of the right-hand set have an incoming arrow. The map is onto even though two inputs share output 1, so onto does not imply one-to-one.

Method

For $f:A\to B$ , surjectivity means: for every $y\in B$ , there exists $x\in A$ with $f(x)=y$ . Verify both that the constructed input belongs to $A$ and that substitution returns $y$ .

Worked example

Take $f:\mathbb R\to\mathbb R$ , $f(x)=3x-2$ . Given any real $y$ , set $x=(y+2)/3$ . This is real and $f(x)=3(y+2)/3-2=y$ . Therefore the function is onto. On domain $[0,\infty)$ with the same codomain, it is not onto: its range starts at $-2$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Which choice makes f(x)=x² onto?

Domain ℝ, codomain ℝ

Domain (0,∞), codomain [0,∞)

Domain ℝ, codomain [0,∞)

Hint 1 · Find a starting point

Every value in the codomain must be reached.

Hint 2 · Take the next step

Check negative outputs and whether zero is attained.

Show the reasoning

Answer: Domain ℝ, codomain [0,∞)

With all real inputs, every nonnegative output occurs, including zero.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Common mistake

Checking a few targets is evidence, not a proof for every target. Use a symbolic arbitrary $y$ .

Check your understanding

Is $x^3:\mathbb R\to\mathbb R$ onto?

Show answer

Yes. Every real $y$ has the real preimage $\sqrt[3]y$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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