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I got this problem from Munkres, my idea is similar, but comparing to the actual solution, I missed at least 4 steps.

Prove that the projection maps $\pi_1 : X \times Y \to X$ and $\pi_2 : X \times Y \to Y$ are both open maps.

Here is my solution

Let $U_1 \times U_2 \subset X \times Y$ be open in $X \times Y$, meaning $U_1$ and $U_2$ are open in $X$ and $Y$ respectively that is $U_1 \in \tau_X$ and $U_2 \in \tau_Y$. Now by the definition of $\pi_1$, we have $$\pi_1(U_1 \times U_2) = U_1 \in \tau_X$$ so the image is also open. The proof for $\pi_2$ is similar.

Now the solution in page 4 of the pdf includes all these union and basis things. The only major thing I noticed is that I didn't consider either $U_1$ or $U_2$ to be empty, but I am not sure how meaningful it is to project on empty sets because the projection is still empty and that is clearly open.

Lemon
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2 Answers2

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I think the main issue with your solution is that you consider only those open sets in the product topology that are of the form $U_1\times U_2$ for some $U_1$ and $U_2$ open in $X$ and $Y$, respectively. Sure, sets of this form are open in the product topology, but there are many more open sets that cannot be expressed as simple products.

In particular, $V\subseteq X\times Y$ is open in the product topology if and only if there exist collections of open sets $\{U_1^{\alpha}\}_{\alpha\in A}$ in $X$ and $\{U_2^{\alpha}\}_{\alpha\in A}$ in $Y$, where $A$ is a non-empty index set, such that $$V=\bigcup_{\alpha\in A}U_1^{\alpha}\times U_2^{\alpha}.$$

As for the issue with empty sets, if either $U_1$ or $U_2$ is empty, then $U_1\times U_2$ is empty and the image of the empty set is vacuously empty.

triple_sec
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  • I am still confused, what is the introduction of the collections of open sets and unions for?? – Lemon Jun 06 '14 at 02:49
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    @sidht Be careful: the product topology on $X\times Y$ is not ${U_1\times U_2,|,U_1\in\tau_X,U_2\in\tau_Y}$, but the collection of all unions of sets of this form! This comes from the very definition of the product topology. – triple_sec Jun 06 '14 at 02:53
  • Here is a nice example of why sheer products are not enough and why unions thereof are needed. – triple_sec Jun 06 '14 at 02:54
  • @triple_sec, what if we change the topology to open rectangles instead of open balls? Then it would work. – Lemon Jun 06 '14 at 02:59
  • @sidht I'm afraid it wouldn't work, either. The union of two rectangles need not be a rectangle, but a topology should be closed under unions, by definition. I believe it would be very helpful for you to consult the definition of the product topology. It's §15 in Munkres (pp. 86–88). – triple_sec Jun 06 '14 at 03:04
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    Actually, I think this should be fine - unless I'm mistaken, you can check that $f : X\to Y$ is open by checking on a basis, which is what sidht is doing. – Stahl Jun 06 '14 at 03:53
  • @triple_sec, you are right, that was hasty of me. Union of two disjoint rectangles is not a rectangle. – Lemon Jun 06 '14 at 04:44
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    @Stahl You're right; the image mapping does happen to commute with unions (but not, in general, with intersections and complements, unlike the inverse image). I did feel, however, some confusion about realizing that the product topology is more than just the collection of products of open sets. Ultimately, what sidht did is indeed sufficient to see the openness of the projection maps, but explicit reference ought to be made how the result for the topological basis implies the result for the whole topology, as you mentioned. You might want to point this out in the form of an answer. – triple_sec Jun 06 '14 at 05:20
  • @triple_sec, thanks you did raised a point I was not aware of. – Lemon Jun 07 '14 at 05:28
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    Both answers have been very helpful, thank you. – Lemon Jun 13 '14 at 04:31
  • @sidht My pleasure, it was an interesting problem. – triple_sec Jun 13 '14 at 22:56
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As I noted in the comments of triple_sec's answer, what you did is in fact sufficient to prove that projections are open mappings. The reason your proof is enough is that a basis for the product topology on $X\times Y$ is given by sets of the form $U_1\times U_2$, with $U_1\subseteq X$ open and $U_2\subseteq Y$ open. In general, it is true that $f(A\cup B) = f(A)\cup f(B)$, so it is sufficient to check openness of a map $f : X\to Y$ on a basis for $X$ (i.e., check that any basis open has open image in $Y$ under $f$), the same way that it is sufficient to check continuity of $f$ on a basis of $Y$ (check that any basis open in $Y$ has open preimage in $X$). Of course, if you are proving statements like "a projection map is open," you will probably be expected to say why it's OK that you only check openness on a basis.

However, one does need to be careful, as a basis $\beta$ for a topology $\tau$ on $X$ is not the whole story: a general open set in $\tau$ is some union of basis opens, but not necessarily an element of $\beta$ itself! Although there are a number of properties that can be checked on a basis, there are also properties that cannot be checked on a basis, so one must realize the difference between $\beta$ and $\tau$ when examining a space $X$.

Stahl
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  • Thanks, but what did you mean in the 2nd paragraph when you say "is some union of the basis, but not necessarily an element of the basis itself!". Doesn't this contradict basis being closed under union? – Lemon Jun 06 '14 at 06:13
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    @sidht I don't know of any definition of basis that requires it to be closed under unions. In fact, this is kind of the point of a basis: it is essentially a set of "simple" open sets such that any open set in the topology is a union of such simple opens (opens in the basis). It is the topology itself that needs to be closed under unions. – Stahl Jun 06 '14 at 06:55
  • A collection of subsets of $X$, $\mathscr B$, is a (topological) basis if $\mathscr B\subseteq\mathscr\tau$ and every non-empty member of $\tau$ can be expressed as a union of sets in $\mathscr B$. Note that a basis $\mathscr B$ need not be closed under unions, as opposed to a topology $\tau$, which, of course, is closed under unions. It is possible that a base is much smaller than the topology it defines. Indeed, there exist topologies with uncountably many sets, yet having a countable basis. These topological spaces are called second countable. E.g., $\mathbb R^n$ is second countable. – triple_sec Jun 06 '14 at 08:10
  • @Stahl, you are correct. My mistake, I was thinking about something else (not the topology). – Lemon Jun 07 '14 at 05:28
  • Also, I have to confess that I was not aware that not every open set can be decomposed into the cross product as triple_sec has indicated. This proof of mine is more of an accident from a transition of Rudin's analysis to this. – Lemon Jun 07 '14 at 05:29
  • So just to clarify, basically I used $$f(W) = f(\cup_{i \in I} U_i \times V_i) = \cup_{i \in I} f(U_i \times V_i)$$ without stating that (1) finite/arbitrary union of open sets preserves openness (2) elements in the topology is union of elements in base, so that $U_1 = \cup_{\alpha} U_1^\alpha$ in the notation triple_sec used? – Lemon Jun 07 '14 at 05:32
  • Your first formula is correct. I.e., if $X$ is an arbitrary space, $I$ is a nonempty index set, $U_i\subseteq X$ for all $i\in I$, and $f:X\to Y$ is a function, then $f(\cup_{i\in I}U_i)=\cup_{i\in I}f(U_i)$. Proof: if $y\in f(\cup_{i\in I}U_i)$, then $y=f(x)$ for some $x\in\cup_{i\in I}U_i$, i.e., $\exists j\in I$ such that $x\in U_j$, so $y=f(x)\in f(U_j)\subseteq f(\cup_{i\in I}U_i)$. The other direction is similarly easy. Exercise: show that this formula does not hold for intersections or complements, in general. – triple_sec Jun 07 '14 at 05:41
  • As for $U_1=\cup_{\alpha}U_1^{\alpha}$, be careful, for in general, it is not true that $(\cup_{\alpha\in A}U_1^{\alpha})\times(\cup_{\alpha\in A}U_2^{\alpha})=\cup_{\alpha\in A}U_1^{\alpha}\times U_2^{\alpha}$. Check why it fails for $U_1^{\alpha}=U_2^{\alpha}=[1,3]$ and $U_1^{\beta}=U_2^{\beta}=[2,4]$. – triple_sec Jun 07 '14 at 05:44
  • That is, if $V$ is open in the product topology, then it can be expressed as $V=\cup_{\alpha\in A}U_1^{\alpha}\times U_2^{\alpha}$ (as argued before), but, defining $U_1\equiv \cup_{\alpha\in A}U_1^{\alpha}$ and $U_2\equiv \cup_{\alpha\in A}U_2^{\alpha}$, it is not necessarily true that $V=U_1\times U_2$! Again, not every open set in the product topology can be expressed as a simple product of open sets. – triple_sec Jun 07 '14 at 05:50
  • Both answers have been very helpful, thank you. – Lemon Jun 13 '14 at 04:32