4
$\begingroup$

Let $X$ be a topological space and $f: X^2\to X$ be a projection onto the first factor.

Is $f$ continuous?

Thanks for your help.

$\endgroup$
0
13
$\begingroup$

The continuity of $f$ strongly relies (of course) on the topology $X^{2}$ is provided with. If $X^{2}$ has the product topology, then the projections onto the factors are clearly continuous (if you like, this is just by definition of product topology, see here).

$\endgroup$
5
  • 1
    $\begingroup$ So what is the answer? $\endgroup$
    – hengxin
    Sep 25 '14 at 13:31
  • $\begingroup$ @hengxin Well, the answer is that, IF $X^2$ is endowed with the product topology (as it is usually understood to be), then $f$ is continuous. But, IF $X^2$ has an arbitrary topology, then the answer is that, in general, $f$ is not continuous. $\endgroup$ Sep 25 '14 at 14:05
  • $\begingroup$ Thanks. As a beginner of general topology, I am still often confused with so much definitions, and need some definite answers. $\endgroup$
    – hengxin
    Sep 25 '14 at 14:09
  • 1
    $\begingroup$ @hengxin You are welcome! Anyway, as I more or less underlined above, if you find in a book sentences like: "Let $X$ and $Y$ be topological spaces and let $X\times Y$ be their product", then you can assume that $X\times Y$ is intended to denote the product of $X$ and $Y$ as topological spaces, that is the cartesian product of the underlying sets of $X$ and $Y$ endowed with the product topology. If this had not to be the case, the author would explicitly say so. $\endgroup$ Sep 25 '14 at 14:21
  • $\begingroup$ I have another question about the product space. Would you mind checking it out at your convenience? $\endgroup$
    – hengxin
    Sep 26 '14 at 1:11
7
$\begingroup$

If $\{X_\omega\}_{\omega \in \Omega}$ is a family of topological spaces, it is standard practice to turn the corresponding Cartesian product $\prod_{\omega \in \Omega}X_\omega$ into a topological space as well by equipping it with the so-called "product topology."

Now, the important thing to realize is that this product topology is expressly defined so as to make all the projections $\pi_\alpha:\left( \prod_{\omega \in \Omega}X_\omega \right) \rightarrow X_\alpha$ continuous.1

Therefore, absent any specific information to the contrary, I'd say that the answer to your question is yes.

(But, as already mentioned, the question of continuity does depend critically on the topologies chosen—after all, it is the topologies of the domain and codomain that define which functions are continuous. This means that, if, contrary to standard practice, the topology assigned to your $X^2$ is not the standard product topology described above, then all bets are off.)


1And in fact, this product topology is defined to be the smallest (aka weakest) topology with this property. More specifically, the product topology is defined as the topology on $\prod_{\omega \in \Omega}X_\omega$ that is generated by the subbase

$$\bigcup_{\omega \in \Omega}\{\pi_\omega^{-1}(U):U\text{ is open in } X_\omega\}$$

This means that the product topology is the smallest topology on the product space that contains all the inverse images of open sets with respect to some projection $\pi_\omega$. Therefore, in this topology, the projections $\pi_\omega$ are rendered continuous by construction.

$\endgroup$
1
  • 1
    $\begingroup$ Universal property for the win! $\endgroup$ May 8 '13 at 9:20
6
$\begingroup$

If $U$ is open in $X$, then $U\times X$ is open in $X^2$.

$\endgroup$
2
  • $\begingroup$ So what is the answer? $\endgroup$
    – hengxin
    Sep 25 '14 at 13:31
  • $\begingroup$ The preimage of every open set is an open set, which is the definition of continuity. $\endgroup$
    – RShields
    Feb 10 '20 at 17:35
0
$\begingroup$

A known example where projections are not continuous in general is the space of cadlag functions on $[0, 1], $ $ D[0, 1]. $ If this space is equipped with the Skorokhod topology (also known as $J_1 $) and we define the projection as $\pi_x: f \mapsto f(x), $ then the projection is continuous if and only if $x $ is a point of continuity for $f. $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.