Composition of Cartesian Products of Mappings

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: