Let $X_1, X_2$ be topological spaces and $X_1 \times X_2$ with the product topology. We define the projection map $\Pi_i : X_1 \times X_2 \rightarrow X_i, \Pi_i(x_1,x_2) = x_i$. Consider the following proposition:

The product topology on $X_1 \times X_2$ is coarsest such that $\Pi_1, \Pi_2$ are continuous.

What is this proposition saying? I've read the proof (which goes on to show that for some $U_i$ open in $X_i$, we have that $U_1 \times U_2$ is open in $X_1 \times X_2$) but I don't understand what the aims of it are or what we're even trying to show.


2 Answers 2


Great question! I don't actually like this characterization of the product of two spaces all that much; to me, it seems to be hiding the real reason we are interested in this space, which is that it is the category-theoretic product of these spaces.

What does this mean? Well, first of all, ask yourself a question about sets, rather than topological spaces. If $A,B,C$ are sets, what does it mean to give a function from $C\to A\times B$?

After a bit of thought, you realize that specifying a function from $C\to A\times B$ is exactly the same as specifying a pair of functions $f\colon C\to A$ and $g\colon C\to B$. We may write the function as $(f,g)\colon C\to A\times B$, sending $c$ to $(f(c),g(c))$.

If we like, we can recover the functions $f$ and $g$ from the funciton $(f,g)$ by composing it with the projection maps $\Pi_1\colon A\times B\to A$ and $\Pi_2\colon A\times B\to B$. Put another way, if $h\colon C\to A\times B$ is any function, then we always have $h=(\Pi_1\circ h, \Pi_2\circ h)$.

The idea for the product of too topological spaces is the same, except now we replace 'set' with 'topological space' and 'function' with 'continuous function'.

Proposition If $X,Y,Z$ are topological spaces and $h\colon Z\to X\times Y$ is continuous, then $\Pi_1\colon X\times Y\to X$ and $\Pi_2\colon X\times Y\to Y$ are continuous and $h=(\Pi_1\circ h, \Pi_2\circ h)$.

In other words, continuous functions into a product space correspond precisely to pairs of continuous functions going into each of the two spaces.

What has this to do with your proposition? Well, in category theory, we have something called a limit, and a product is a particular special case of a limit. A beautiful fact about topological spaces is that we may always construct limits by taking the appropriate limit in the category of sets, and then endowing this set with the coarsest possible topology such that the appropriate functions are continuous.

If you don't know category theory, that paragraph probably didn't make much sense, so I'll try to illustrate it in this case.

We start with two topological spaces $X$ and $Y$ and we want to construct the product space so that the proposition above is satisfied. First, we pass to the underlying sets $X$ and $Y$ of these topological spaces. We already know how to form the product of these sets - it is just the set $X\times Y$ of pair $(x,y)$. Now we want to know what topology to put on this set.

Firstly, we are going to need $\Pi_1$ and $\Pi_2$ to be continuous. This means that $\Pi_1\circ h$ and $\Pi_2\circ h$ are automatically going to be continuous. In order to make the proposition hold, we need the maps $(f,g)$ to be continuous whenever $f,g$ are continuous.

The coarser the topology we put on $X\times Y$, the 'easier' it will be for a function into $X\times Y$ to be continuous, since there will be fewer open sets to worry about. And it turns out that if we take the coarsest topology of all (subject to $\Pi_1,\Pi_2$ being continuous) then the proposition holds.

  • $\begingroup$ This is a great answer, thank you for taking the time to explain the underlying relationship in such depth. This is an answer that I'll definitely be reading again as I progress through this course, and courses over the next few years. Most of this has passed over my head the first time through and required a couple of rereads in some parts. $\endgroup$ May 19, 2016 at 12:24
  • $\begingroup$ @IrregularUser If you tell me which parts are unclear, I can try to rewrite them. $\endgroup$ May 19, 2016 at 13:23
  • $\begingroup$ Great answer, but I have one objection: we are interested in the product construction in topology because it generalizes many important examples, starting with $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$. $\endgroup$
    – Lee Mosher
    May 19, 2016 at 13:25

Intuitively: in order for the projection map to be continuous, we need there to be "lots" of open sets in $X_1 \times X_2$ in order for $f^{-1}(U_i)$ to be open for all open $U_i$ in $X_i$. The more open sets we have in $X_1 \times X_2$ the greater chance there is that $f^{-1}(U_i)$ is always open. The proposition says that the product topology has the fewest open sets necessary: so any other topology for which $f^{-1}(U_i)$ is always open must contain all the open sets of the product topology.

  • $\begingroup$ Oh, I think I kind of get it now! Then how does the proof show this? It takes any open $U_i \subset X_i$ and shows that $U_1 \times U_2$ is open in the product topology. How do we know it's not open in some smaller topology? $\endgroup$ May 19, 2016 at 11:43
  • $\begingroup$ You show that for any topology $T$ on $X_1 \times X_2$ that the product topology must be a subset of $T$. You do that in the usual way for set: by taking an arbitrary $U$ open in the product topology, and showing $U$ must also be open in topology $T$. $\endgroup$
    – Josh R
    May 19, 2016 at 12:06

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