Composition of Mappings is Associative

From ProofWiki
Jump to: navigation, search

Theorem

The composition of mappings is an associative binary operation:

$\paren {f_3 \circ f_2} \circ f_1 = f_3 \circ \paren {f_2 \circ f_1}$

where $f_1, f_2, f_3$ are arbitrary mappings which fulfil the conditions for the relevant compositions to be defined.


Proof

From the definition, we know that a mapping is a relation.

First, note that from the definition of composition of relations, the following must be the case before the above expression is even to be defined:

$(1): \quad \Dom {f_2} = \Cdm {f_1}$
$(2): \quad \Dom {f_3} = \Cdm {f_2}$

where $\Cdm f$ denotes the codomain of the mapping $f$.


The two composite relations can be seen to have the same domain, as follows:

\(\displaystyle \Dom {\paren {f_3 \circ f_2} \circ f_1}\) \(=\) \(\displaystyle \Dom {f_1}\) $\quad$ Domain of Composite Relation $\quad$


\(\displaystyle \Dom {f_3 \circ \paren {f_2 \circ f_1} }\) \(=\) \(\displaystyle \Dom {f_2 \circ f_1}\) $\quad$ Domain of Composite Relation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \Dom {f_1}\) $\quad$ Domain of Composite Relation $\quad$


Also they have the same codomain, as is seen by:

\(\displaystyle \Cdm {\paren {f_3 \circ f_2} \circ f_1}\) \(=\) \(\displaystyle \Cdm {f_3 \circ f_2}\) $\quad$ Codomain of Composite Relation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \Cdm {f_3}\) $\quad$ Codomain of Composite Relation $\quad$


\(\displaystyle \Cdm {f_3 \circ \paren {f_2 \circ f_1} }\) \(=\) \(\displaystyle \Cdm {f_3}\) $\quad$ Codomain of Composite Relation $\quad$


As a mapping is a relation, we can use that the Composition of Relations is Associative‎:

$\forall x \in \Dom {f_1}: \map {\paren {f_3 \circ f_2} \circ f_1} x = \map {f_3 \circ \paren {f_2 \circ f_1} } x$

Hence the result.

$\blacksquare$


Sources