Continuous Mapping to Product Space/General Result

Theorem
Let $X$ be a topological space.

Let $\mathbb S = \sequence {Y_i}_{i \mathop \in I}$ be a sequence of topological spaces for some indexing set $I$.

Let $\displaystyle Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\mathbb S$.

Let $f$ be a mapping from $X$ to $Y$.

Then $f$ is continuous $\pr_i \circ f$ is continuous for all $i \in I$.

Proof
Suppose $f$ is continuous.

From Projection from Product Topology is Continuous $\pr_i$ is continuous for each $i \in I$.

By Continuity of Composite Mapping it follows that $\pr_i \circ f$ is continuous for each $i \in I$.

Conversely, suppose that each $\pr_i \circ f$ is continuous.

Let $U = \pr_{i_1}^{-1} \sqbrk {U_{i_1} } \cap \cdots \cap \pr_{i_n}^{-1} \sqbrk {U_{i_n} }$ be an open set in the natural basis of the product topology for $Y$.

Note that by the definition of natural basis, $U$ is constrained to be the intersection of a finite number of sets of $\mathbb S$.

Then by Preimage of Intersection under Mapping we find:


 * $f^{-1} \sqbrk U = f^{-1} \sqbrk {\pr_{i_1}^{-1} \sqbrk {U_{i_1} } } \cap \cdots \cap f^{-1} \sqbrk {\pr_{i_n}^{-1} \sqbrk {U_{i_n} } }$

Since each $\pr_{i_j} \circ f$ is continuous, each:


 * $\paren {\pr_{i_j} \circ f}^{-1} \sqbrk {U_{i_j} } = f^{-1} \sqbrk {\pr_{i_j}^{-1} \sqbrk {U_{i_j} } }$

is open.

Therefore $f^{-1} \sqbrk U$ is open.

By the definition of basis, each open set of $Y$ can be written as a union of sets in the natural basis of the product topology for $Y$.

From Preimage of Union under Mapping, the union of the preimages of open sets of $Y$ equals the preimages of the union of those sets.

Thus the preimage of an open set of $Y$ is open in $X$.

Hence $f$ is continuous.