Function to Product Space is Continuous iff Composition with Projections are Continuous

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Theorem

Let $X$ be a topological space.

Let $\mathbb S = \left\langle {Y_i}\right \rangle_{i \in I}$ 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 if and only if $\operatorname{pr}_i \circ f$ is continuous for all $i \in I$.


Proof

Suppose $f$ is continuous.

From Projection from Product Topology is Continuous $\operatorname{pr}_i$ is continuous for each $i \in I$.

By Continuity of Composite Mapping it follows that $\operatorname{pr}_i \circ f$ is continuous for each $i \in I$.


Conversely, suppose that each $\operatorname{pr}_i \circ f$ is continuous.

Let $U = \operatorname{pr}_{i_1}^{-1} \left({U_{i_1}}\right) \cap \cdots \cap \operatorname{pr}_{i_n}^{-1} \left({U_{i_n}}\right)$ 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:

$f^{-1} \left({U}\right) = f^{-1} \left({\operatorname{pr}_{i_1}^{-1} \left({U_{i_1}}\right)}\right) \cap \cdots \cap f^{-1} \left({\operatorname{pr}_{i_n}^{-1} \left({U_{i_n}}\right)}\right)$

Since each $\operatorname{pr}_{i_j} \circ f$ is continuous, each:

$\left({\operatorname{pr}_{i_j} \circ f}\right)^{-1} \left({U_{i_j}}\right) = f^{-1} \left({\operatorname{pr}_{i_j}^{-1} \left({U_{i_j}}\right)}\right)$

is open.

Therefore $f^{-1} \left({U}\right)$ 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.

$\blacksquare$