Composition of Cartesian Products of Mappings

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Theorem

Let $I$ be an indexing set.

Let $\family {S_\alpha}_{\alpha \mathop \in I}$, $\family {T_\alpha}_{\alpha \mathop \in I}$ and $\family {U_\alpha}_{\alpha \mathop \in I}$ be families of sets all indexed by $I$.


For each $\alpha \in I$, let:

$f_\alpha: S_\alpha \to T_\alpha$ be a mapping
$g_\alpha: T_\alpha \to U_\alpha$ be a mapping.


Let:

$\ds S = \prod_{\alpha \mathop \in I} S_\alpha$
$\ds T = \prod_{\alpha \mathop \in I} T_\alpha$
$\ds U = \prod_{\alpha \mathop \in I} U_\alpha$


Let $f: S \to T$ and $\ds g: T \to U$ be defined as:

$\ds f = \prod_{\alpha \mathop \in I} f_\alpha$
$\ds g = \prod_{\alpha \mathop \in I} g_\alpha$


Then their composition $g \circ f: S \to U$ is:

$\ds g \circ f: \prod_{\alpha \mathop \in I} g_\alpha \circ f_\alpha$


Proof

First note that for all $\alpha \in I$:

$\Dom {g_\alpha} = \Cdm {f_\alpha} = T_\alpha$

where $\Dom {g_\alpha}$ denotes the domain of $g_\alpha$ and $\Cdm {f_\alpha}$ denotes the codomain of $f_\alpha$.

So $g_\alpha \circ f_\alpha$ is defined for all $\alpha \in I$.


Similarly:

$\Cdm f = \Dom g = T$

and so $g \circ f$ is defined.


Let $\mathbf x \in S$ be arbitrary:

$\mathbf x = \family {x_\alpha \in S_\alpha}_{\alpha \mathop \in I}$

We have that:

\(\ds \map {\paren {g \circ f} } {\mathbf x}\) \(=\) \(\ds \map {\prod_{\alpha \mathop \in I} g_\alpha} {\map {\prod_{\alpha \mathop \in I} f_\alpha} {\mathbf x} }\) Definition of $f$
\(\ds \) \(=\) \(\ds \map {\prod_{\alpha \mathop \in I} g_\alpha} {\family {\map {f_\alpha} {x_\alpha} }_{\alpha \mathop \in I} }\) Definition of $\ds \prod_{\alpha \mathop \in I} f_\alpha$
\(\ds \) \(=\) \(\ds \family {\map {\paren {g_\alpha \circ f_\alpha} } {x_\alpha} }_{\alpha \mathop \in I}\)
\(\ds \) \(=\) \(\ds \map {\prod_{\alpha \mathop \in I} g_\alpha \circ f_\alpha} {\mathbf x}\)

$\blacksquare$


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