# Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal

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## Theorem

Let $A$ and $B$ be classes.

Let $f$ and $g$ be mappings on $A \times B$.

Let $f$ and $g$ be such that:

- $f \circ g^{-1} = I_B$

where:

- $g^{-1}$ denotes the inverse of $g$
- $I_B$ denotes the identity mapping on $B$
- $\circ$ denotes composition of mappings.

Then:

- $f = g$

The validity of the material on this page is questionable.In particular: Each of the definitions inverse, identity mapping and composition of mappings is defined on sets only. They need to be expanded to classes.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Questionable}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Proof

Let $\tuple {a, b} \in f$.

Then because $\tuple {b, b} \in I_B$ we must have:

- $\tuple {b, a} \in g^{-1}$

and so by definition of inverse of mapping:

- $\tuple {a, b} \in g$

Hence:

- $f \subseteq g$

Let $\tuple {a, b} \in g$.

Then by definition of inverse of mapping:

- $\tuple {b, a} \in g^{-1}$

Then because $\tuple {b, b} \in I_B$ we must have:

- $\tuple {a, b} \in f$

Hence:

- $g \subseteq f$

Hence by definition of set equality:

- $f = g$

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries