Equality of Mappings

Theorem

Two mappings $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal iff:

• $S_1 = S_2$
• $T_1 = T_2$
• $\forall x \in S_1: f_1 \left({x}\right) = f_2 \left({x}\right)$.

Proof

This follows directly from Equality of Relations.

$\blacksquare$

Comment

It is worth labouring the point that for two mappings to be equal, not only must their domains be equal, but so must their codomains.

However, note that some sources, for example W.E. Deskins: Abstract Algebra (1964), do not impose this condition, stating merely that two mappings are equal if the domains are equal and every element has the same image under each mapping.

Other sources, for example Ian D. Macdonald: The Theory of Groups (1968), gloss over the concepts of domain and codomain and merely state the equality of the images.

Sources

• Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts (1951): Introduction $\S 2$
• Iain T. Adamson: Introduction to Field Theory (1964)... (previous)... (next): $\S 1.3$
• W.E. Deskins: Abstract Algebra (1964): $\S 1.3$: Theorem $1.12$
• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 3.2$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 1$
• Ian D. Macdonald: The Theory of Groups (1968): Appendix
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 10$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 4$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 20 \ \text {(j)}$
• John F. Humphreys: A Course in Group Theory (1996): $\S 2$