Proof: Categorical Product = Topological Product

Question

Is there a nice way to prove that the categorical product equals the topological product?

What I mean is the following:

Starting with a given family of topological spaces $X_i$ and any topological space $Z$ together with continuous mappings $f_i:Z\to X_i$ let me construct a categorical product in the category $\mathbf{Top}$.

First, I consider them as merely sets so I can construct a categorical product in the category $\mathbf{Set}$.
That is taking the cartesian product $\prod_{i\in I}X_i$ as object and the cartesian projections $\pi_i:\prod_{i\in I}X_i\to X_i$ as morphisms there is a uniquely determined morphism $f:Z\to \prod_{i\in I}X_i$ s.t. the diagram commutes $f\circ\pi_i=f_i$.

Now, I start to equip that product with some topology and I have to do that in such a way that the projections $\pi_i$ are continuous and that the respective mappings $f$ are continuous as well; so I see that the product topology must neither be to fine nor to coarse.
Since it has to hold for arbitrary objects $Z$ I rather start with the projections $\pi_i$ than with the mappings $f$; thus, I choose the coarsest topology s.t. the projections $\pi_i$ are still continuous and hope that this topology is coarse enough to make the mappings $f$ continuous.
That is I take all sets $\pi_i^{-1}(U_i)$ as generator for my product topology. That is certainly the coarsest topology s.t. the projections $\pi_i$ are still continuous.

At that point, I'm stuck by showing that the mapping $f$ actually become continuous w.r.t. the chosen topology!

Finally, this uniquely defines a categorical product in the category $\mathbf{Top}$ up to isomorphism by category theoretic considerations - so I'm done from the category theory side.

Next, I can check wether this is actually the topological product - but this is a rather easy exercise:
Given any map $f:Z\to\prod_{i\in I}X_i$.
If $f$ is continuous so is $\pi_i\circ f$ continuous as composition of continuous maps.
Conversely, if $\pi_i\circ f$ is continuous so is $f$ continuous by the existence and uniqueness of morphism for a categorical product.

Besides I want to understand:

Apart from the fact that the categorical product is unique up to isomorphism the question now arises wether the topological product is unique since careful considerations show two aspects:

1. The start off for a topological product is different data:
   $X_i$, $X$(new!) plus $\pi_i:X\to X_i$(new!) and $Z$ together with $f:Z\to X$(switched! + merely map!) where $X:=\prod_{i\in I}X_i$
2. The characteristic property does not claim the existence and uniqueness of an according morphism making the diagram commute:
   $f_i:=\pi_i\circ f$(taken not given!) and $f$(given not stated!)

So, going from the characteristic(!) property (I strictly distinguish from universal property here) I cannot claim the uniqueness of the topological product up to isomorphism without referring to the relation to the categorical product and its universal(!) property.

So, does somebody know how to carefully!!!!!! untwine this - again I want to stress carefully!!!!!! and not by some pretty standard blah blah blah which I read already here and there (I'm really sorry for being so rude in here - I just want to pretend by sort of everytime popping out nowhere leading answers)

Thanks everybody alot for hints, insights and ideas! =)

Answer 1

So if I understand correctly, what you want is a proof of the

Theorem: Let $\left\{ f_i \colon X \to Y_i \mid i \in I\right\}$ a family of maps, where the $Y_i$ are topological spaces, and $X$ is a set. If $\tau_1$ and $\tau_2$ are topologies on $X$ with the property that a map $g \colon (Z,\tau_Z) \to (X,\tau_k)$ is continuous if and only if $f_i \circ g \colon (Z,\tau_Z) \to (Y_i,\tau_{Y_i})$ is continuous for all $i\in I$, then $\tau_1 = \tau_2$.

Proof: Since $\operatorname{id} \colon (X,\tau_k) \to (X,\tau_k)$ is continuous, it follows that $f_i \colon (X,\tau_k) \to (Y_i,\tau_{Y_i})$ is continuous for all $i\in I$, for $k \in \{1,2\}$. Choosing $(X,\tau_2)$ for $(Z,\tau_Z)$ and $g = \operatorname{id}$ in the universal property for $\tau_1$, we find that $g \colon (X,\tau_2) \to (X,\tau_1)$ is continuous, since $f_i \circ g = f_i \colon (X,\tau_2) \to (Y_i,\tau_{Y_i})$ is continuous for all $i\in I$ by the above. Hence $\tau_1 \subset \tau_2$. Swapping the roles, we obtain $\tau_2 \subset \tau_1$, and thus $\tau_1 = \tau_2$.

That shows that the universal property characterises the initial topology uniquely - if it exists.

It remains to see that a topology with the universal property exists. If such a topology exists, it must be the coarsest topology with respect to which all $f_i$ are continuous, hence it must be the topology $\tau$ that has

$$\mathcal{S} = \bigcup_{i\in I}\left\{ f_i^{-1}(U) : U \in \tau_{Y_i}\right\}$$

as a subbasis.

We must see that $\tau$ has the universal property. We note that a map $g \colon (Z,\tau_Z) \to (X,\tau)$ is continuous if and only if $g^{-1}(S)$ is open for all $S \in \mathcal{S}$ (every open set is a union of finite intersections of such sets, hence the preimage of an open set is then a union of finite intersections of open sets, which is open). If $g$ is continuous, then $f_i \circ g$ is a composition of continuous maps, hence continuous, for all $i\in I$. Conversely, if $f_i \circ g$ is continuous for all $i\in I$, and $S\in \mathcal{S}$, say $S = f_{i_0}^{-1}(U_0)$ for an open $U_0 \subset Y_{i_0}$, then

$$g^{-1}(S) = g^{-1}\left(f_{i_0}^{-1}(U_0)\right) = (f_{i_0}\circ g)^{-1}(U_0)$$

is open. Thus $g$ is continuous, and we have seen that $\tau$ has the universal property, so the existence is established.

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Now it is easy to see that the topological product of spaces $X_i$ - the Cartesian product $X = \prod\limits_{i\in I} X_i$ endowed with the initial topology $\tau$ with respect to the projections $\pi_i \colon X \to X_i$ - is a product in the category $\mathbf{Top}$.

Given a topological space $(Z,\tau_Z)$ and a family of continuous maps $f_i \colon Z \to X_i$, since the Cartesian product is a product in the category $\mathbf{Ens}$ of sets (you may prefer to call it $\mathbf{Set}$, but I had too Bourbakist teachers for that), there is a unique map $f \colon Z \to X$ with $f_i = \pi_i \circ f$ for all $i\in I$. But, by assumption, $f_i = \pi_i\circ f$ is continuous, hence by the universal property of the product topology, $f$ is indeed continuous, i.e. a morphism in $\mathbf{Top}$. The uniqueness follows by applying the forgetful functor $\mathbf{Top}\to\mathbf{Ens}$.

Answer 2

@Besides I want to understand:

I think I got the difference now:

While the universal property defines the object completely, the characteristic property defines the topology only!
That is, there is one and only one(!) topological product $\prod_{i\in I}X_i$ uniquely determined by the universal property - but there can be many(!) pairs $(X,\pi_i)$ just as soon as one chooses a specific pair its topology is uniquely determined by the characteristic property.

For example:

Take the real line $\mathbb{R}$ there is one and only one 'topological product' $\mathbb{R}$ however many different 'subspaces' like $[0,1]$ or $(0,1]$ or $[0,1]\cup[2,3]$.

Moreover:

It can happen that some of those realize a product as well as they happen to be homeomorphic - even though they at first glance seem to satisfy the characteristic property only (e.g. $(0,1)\cong\mathbb{R}$).