# Composite of Injection on Surjection is not necessarily Either

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## Theorem

Let $f$ be an injection.

Let $g$ be a surjection.

Let $f \circ g$ denote the composition of $f$ with $g$.

Then it is not necessarily the case that $f \circ g$ is either a surjection or an injection.

## Proof

Let $X, Y, Z$ be sets defined as:

\(\displaystyle X\) | \(=\) | \(\displaystyle \set {a, b, c}\) | |||||||||||

\(\displaystyle Y\) | \(=\) | \(\displaystyle \set {1, 2}\) | |||||||||||

\(\displaystyle Z\) | \(=\) | \(\displaystyle \set {z, y, z}\) |

Let $g: X \to Y$ be defined in two-row notation as:

- $\dbinom {a \ b \ c } {1 \ 2 \ 2}$

which is seen by inspection to be a surjection.

Let $f: Y \to Z$ be defined in two-row notation as:

- $\dbinom {1 \ 2} {x \ y}$

which is seen by inspection to be an injection.

The composition $f \circ g$ is seen to be:

- $\dbinom {a \ b \ c} {x \ y \ y}$

which is:

- not an injection (both $b$ and $c$ map to $y$)
- not a surjection (nothing maps to $z$).

Hence the result.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $2$: Maps and relations on sets: Exercise $2$