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:
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 $\ds \prod_{\alpha \mathop \in I} f_\alpha$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \family {\map {\paren {g_\alpha \circ f_\alpha} } {x_\alpha} }_{\alpha \mathop \in I}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\prod_{\alpha \mathop \in I} g_\alpha \circ f_\alpha} {\mathbf x}\) |
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $2$