# Composite of Three Mappings in Cycle forming Injections and Surjection

## Theorem

Let $A$, $B$ and $C$ be non-empty sets.

Let $f: A \to B$, $g: B \to C$ and $h: C \to A$ be mappings.

Let the following hold:

- $h \circ g \circ f$ is an injection
- $f \circ h \circ g$ is an injection
- $g \circ f \circ h$ is a surjection.

where:

- $g \circ f$ (and so on) denote composition of mappings.

Then each of $f$, $g$ and $h$ are bijections.

## Proof

First note that from Composition of Mappings is Associative:

- $\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$

and so on.

However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done.

We have that $\paren {h \circ g} \circ f$ and $\paren {f \circ h} \circ g$ are injections.

From Injection if Composite is Injection it follows that:

Similarly, because $h \circ \paren {g \circ f}$ and $f \circ \paren {h \circ g}$ are injections:

We have that $g \circ \paren {f \circ h}$ is a surjection.

From Surjection if Composite is Surjection it follows that $g$ is a surjection.

Similarly, because $\paren {g \circ f} \circ h$ is a surjection:

- $g \circ f$ is a surjection.

Thus we have deduced that:

- $g$ is a bijection

and:

- $g \circ f$ is a bijection.

From Mapping Composed with Bijection forming Bijection is Bijection it follows that:

- $f$ is a bijection.

Thus from Inverse of Bijection is Bijection we have that $f^{-1}: B \to A$ and $g^{-1}: C \to B$ are both bijections.

Similarly, $\paren {g \circ f}^{-1}: C \to A$ is a bijection, which from Inverse of Composite Bijection is the same as $f^{-1} \circ g^{-1}$.

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $16$