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I have some doubts concerning projection mappings and product topologies.

The definition states that given an indexed family of topological spaces $\{ (X_i , \tau_i )_{i \in I} \}$, we can set $X = \prod_{i \in I} X_i $. Hence, $\tau = \prod_{i \in I} \tau_i $ is the product topology on $X$, if every projection mapping $\pi_i : X \to X_i$ is continuous for every $X_i$.

In order to make sense of it, I simply think about the standard topological definition of continuity, i.e. for every $i \in I$, and for every open (resp. closed) $G \subseteq X_i$, the preimage $f^{-1} (G) \subseteq X$ is open (resp. closed).

Questions:

  1. Does a projection mapping that induces the product topology on a family of TS preserve compactness?
    In other words, is the case that for every $i \in I$, and for every compact $G \subseteq X_i$, the preimage $f^{-1} (G) \subseteq X$ is compact?
  2. If it is so, does the property extends to any continuous function, beyond the projection mapping in the product topology case?

Any feedback is more than welcome.
Thank you for your time.

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    $\begingroup$ Consider the product of a circle $S^1$ with the real line $\mathbb{R}$, and look at the projection map onto the circle. Doesn't that give a counter-example? $\endgroup$ – mathmandan Mar 18 '15 at 16:23
  • $\begingroup$ @mathmandan: Thanks a lot for the comment. May I ask you to be more explicit (maybe with an answer, I will be happy to accept)? Indeed, I am not completely sure I completely see the counterexample, because I don't really see why the subset of $S^1$ is not compact. Indeed, if we take $[0,1] \in \mathbb{R}$, that induces two "sublines" on $S^1$, but I don't see why they are not compact. $\endgroup$ – Kolmin Mar 18 '15 at 16:29
  • $\begingroup$ Sudden thought: Are they not compact because they are disconnected? (I know it can all sound trivial to you, but my basics really fail from time to time) $\endgroup$ – Kolmin Mar 18 '15 at 16:32
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First of all, the topology $\prod_i\tau_i$ is not the product topology, but the so-called box topology. They only coincide if the familly is finite. The true product topology is generated by products of open sets $U_i\in\tau_i$ (up to here, this is the box topology), but those open sets must be $U_i=X_i$ except for finitely many $i$'s.

Second, the product topology makes all projections continuous, but this does not characterize it. The property that characterizes it is that a mapping $\ f:Y\to\prod_iX_i$ is continuous if and only if all components $f_i=\pi_i\circ f$ are continuous (note the if and only if).

Next, concerning the questions about inverse images of compact sets, this is essentially related to the notion of proper map. In absolute generality proper means universally closed, that is the mapping is closed and any extension $f\times$Id${}_Y$ is closed too (look at wiki, for instance). However technical this is, under mild restrictions this is equivalent to what you asked for: inverse images of compact sets are compact. Now given $\pi_i:X_i\times Y_i\to X_i$ (denote $Y_i=\prod_{j\ne i}X_j$) and $K\subset X_i$ compact, $\pi_i^{-1}(K)=K\times Y_i$ is compact if and only if (Thychonoff) $Y_i$ is compact, if and only (Tych again) if all $X_j$ are compact. Thus projections are proper if and only if all factors of the product are compact.

Finally suppose the factors $X_i$ compact and also Hausdorff. Consider any continuous mapping $f:X\to X_i$. Then any compact set $K\subset X_i$ is closed (by the Hausdorff assumption), hence $f^{-1}(K)$ is closed in $X$, which is compact (Tych once again). But closed in compact is compact, hence $f^{-1}(K)$ is compact. (This is just the general argument to show that a continuous mapping from a compact space to a Haussdorf one is proper.)

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  • $\begingroup$ +1 to your straight to the point answer. Thus, summing up, proper mappings preserve compactness, right? (Thanks a lot for all the other details!) $\endgroup$ – Kolmin Mar 18 '15 at 19:40
  • $\begingroup$ Right, proper maps preserve compactness by inverse image. $\endgroup$ – Jesus RS Mar 18 '15 at 20:26
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Please see the answer of Jesus RS, which includes useful information about the product topology and its relationship to the box topology in general. My answer here is just a requested clarification of an example I suggested as a comment.

Let $X_1$ be the real line $\mathbb{R}$ and $X_2 = S^1$ (under the usual Euclidean metrics). Explicitly, $S^1 = \{z \in \mathbb{C} : \lvert z \rvert = 1\}$. Then let $X = \mathbb{R} \times S^1$ be their Cartesian product, endowed with the product topology (equivalent to the box topology since it's a product of finitely-many spaces, specifically just two spaces).

You can visualize $X$ as a "long cylinder" in 3-dimensional Euclidean space.

Now let $G = \{ 1 \}$ be the subset of $S^1$ that has only a single element, namely the complex number $1 = 1+0i$. Then $G$ is compact (why?), and we look at the projection $\pi_2 : X \to S^1$. The preimage of $G$ under this map is certainly $\mathbb{R} \times G = \mathbb{R} \times \{ 1 \}$, which turns out (under the topology induced as a subspace of the product space $X$) to be homeomorphic to the real line $\mathbb{R}$--see next paragraph if you're not sure about this.

Sketch of proof that $\mathbb{R} \times \{ 1 \}$ is homeomorphic to $\mathbb{R}$: First, the inclusion $\mathbb{R}\times \{ 1 \} \to \mathbb{R} \times S^1$ is continuous (inclusions are always continuous--why?), and you told us that the projection map $\mathbb{R} \times S^1 \to \mathbb{R}$ is continuous, so their composition is a continuous map $f: \mathbb{R}\times \{1 \} \to \mathbb{R}$. This function $f$ is actually quite simple; it's $f(x, 1) = x$. To show that $f$ is a homeomorphism, we'd need to show that $f$ is a bijection (which should now be very simple) and show that $f^{-1}$ is also continuous. This last part basically just requires us to show that for any open set $V \subseteq \mathbb{R}\times \{ 1\}$, the set $f(V)= \{ x : (x,1) \in V \}$ is open in $\mathbb{R}$. $\Box$

But the real line is not compact (show by using the definition of "compact", or recall that compact means closed and bounded in Euclidean space), so you have a counterexample: the preimage of a compact subspace under a projection need not be compact.

Note: it shouldn't be surprising that $\mathbb{R} \times \{ 1 \}$ is homeomorphic to $\mathbb{R}$; indeed it would be shocking if it were anything else. Further, it shouldn't matter if we consider $\mathbb{R} \times \{ 1 \}$ to be a subspace of $\mathbb{R} \times S^1$ or as a product space in its own right...and indeed it doesn't matter at all.

In fact in this example, the circle $S^1$ was irrelevant; you could get the same thing by taking $X_2$ itself to be a single-point space and $G = X_2$, and $X_1$ to be any non-compact space.

This might also be a good time to mention that not all open sets $U$ in the product space $X=X_1 \times X_2$ will be of the form $U_1 \times U_2$ where $U_i$ is open in $X_i$, because you have to also accept any union of such products as open sets. (Think about open sets in $\mathbb{R}$ and open sets in $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$.) Rather, such products $U_1 \times U_2$ form what is called a basis for the box topology. It wasn't quite clear from your question if you had noticed this or not, so I thought I'd throw it in.

Further info:

http://en.wikipedia.org/wiki/Product_topology

http://en.wikipedia.org/wiki/Box_topology

Why are box topology and product topology different on infinite products of topological spaces?

Product and Box Topologies

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  • $\begingroup$ +1. Thanks a lot. Regarding your original comment, when I read $S^1$, I think about a subset of $\mathbb{R}^2$. Hence, I think about two possible projections that map to $\mathbb{R}$. That's why if we take, as I wrote, $[0, 1] \in \mathbb{R}$ we get a disconnected subset of $\mathbb{R}^2$ that corresponds to some "sub"-lines of the original circle. Does what I have written make sense? That's how I actually saw your comment. $\endgroup$ – Kolmin Mar 18 '15 at 19:36
  • $\begingroup$ Moreover, now I don't see another thing regarding your comment. Basing on Jesus' answer, compactness is preserved by continuous projections. Thus, how can your comment constitute a counterexample? Any clarification is much appreciated! :) $\endgroup$ – Kolmin Mar 18 '15 at 19:37
  • $\begingroup$ Regarding your latest comment: a projection may take a non-compact set to a compact set (as in the case of projecting $\mathbb{R}\times \{1 \} \to \{ 1\}$. What it can't do is take a compact set to a non-compact set. proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact $\endgroup$ – mathmandan Mar 18 '15 at 19:41
  • $\begingroup$ Regarding your prior comment, $S^1$ is not the product of two copies of $[0,1]$, as you well know, so while it makes sense to project it onto either the $x$-axis or the $y$-axis, that isn't a projection from a product space. When I said "the product of $\mathbb{R}$ with $S^1$, I really meant $\mathbb{R} \times S^1$, which you may visualize as a subspace of $\mathbb{R}^3$. $\endgroup$ – mathmandan Mar 18 '15 at 19:43
  • $\begingroup$ Just for the sake of fairness, I accepted Jesus questions because it was the first that came, and it was fairly complete in addressing my basic doubts, but again thanks a lot for all the info. $\endgroup$ – Kolmin Mar 19 '15 at 6:45

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