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Product space, $\prod\limits_{a \in I} {X_a}$, where $I$ is from arbitrary index set.

Let $W$ be any topological space,

$f: W \to \prod\limits_{a \in I} {X_a}$ is continuous iff $ \pi_b \circ f: W \to X_b$ for all b (where $\pi_b$ is the projection of $\prod\limits_{a \in I} {X_a}$ into $X_b$) is continuous.

It is claimed that the product topology is the maximal topology such that the equivalence above is true.

What I only know is that product topology is the minimal topology such that projection mapping is continuous. And viewing $\prod\limits_{a \in I} {X_a}$ as the codomain, f would be continuous if indiscrete topology is employed.

So my aim is to "expand" the indiscrete topology, but I don't know how to do it and even after "expansion", why the maximal is the product topology?

Thank you very much!

Martin Sleziak
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1 Answers1

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When $X$ is a set, and $\pi_a:X\to X_a$ are functions from $X$ to spaces $X_a$, indexed by some set $I$, then the initial topology is usually defined as the minimal (or coarsest) topology on $X$ making all $\pi_a$ continuous. Let's call this topology $\tau_i$. If $X=\prod_a X_a$, then the product topology is the initial topology with respect to the projections $\pi_a:\prod_a X_a\to X_a$. It is well-known that $\tau_i$ satisfies the property:

$f: W \to X$ is continuous if $ \pi_a f: W \to X_a$ is continuous for all $a\in I$ (Note the if instead of iff).

Now let $\tau_*$ be any topology on $X$ with this property, and consider the identity function $\text{id}:(X,\tau_i)\to (X,\tau_*)$. Think of $\pi'_a:(X,\tau_*)\to X_a$ as the function $\pi_a$, but this time it is not necessarily continuous. Since $\pi_a\circ\text{id}=\pi_a:(X,\tau_i)\to X_a$ is continuous for each $a\in I$, it follows that $\tau_i$ is finer than $\tau_*$.

If one replaces the if in the property above by an iff, then $\tau_i$ is the unique topology with that property.

Also see Request for gentle explanation of defining a topology with its universal property for the dual situation.

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Stefan Hamcke
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