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!
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:
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.