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Let $X = [0,1]\times[0,1]$ and let $\sim$ be the equivalence relation for the Klein bottle. Let $q: X \to X/\sim$.

Let $C$ be closed in $X$. Then there is closed set $V$ in $\mathbb R^2$ so that $C = V \cap X$.

How to show $q(C)$ is closed in $X/\sim$?

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I would say that it is a more or less direct consequence of the map $q$ being a closed map (quotient maps always are), but I don't know if that's the solution you need. –  Arthur Feb 5 '13 at 14:20
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@Arthur Quotients map do not need to be closed. In this particular instance one way to proceed would be to show the quotient space is Hausdorff and apply the closed map lemma. –  JSchlather Feb 5 '13 at 14:26
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@Arthur: That $q$ is closed is what one wants to prove here, so that's not helpful. Also, projections are quotient maps which are not closed (they are open though). –  Marek Feb 5 '13 at 14:26
    
You're right, both of you. I just checked with my book, and it turns out they are either open or closed. I'm sorry. –  Arthur Feb 5 '13 at 14:27
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2 Answers 2

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If you'd like to show this completely hands-on, here's what you can do. For every closed set $C$ consider set $D := q^{-1}(q(C))$. It is enough to show $D$ is closed because $q(C) = q(D)$ and $q(D)$ is closed precisely when $D$ is (by the definition of quotient topology).

Now, how does $D$ look like? For every point that $C$ shares with the boundary of $X$ you get another point on the opposite (as per the quotient) side. Also, if one of the corners belong to $C$ then all of the corners belong to $D$.

With this picture in mind, let $V$ be a closed set such that $V \cap X = C$. Write $W = V \setminus X^0$. By reflecting, rotating and translating $W$ a few times (in accord with the Klein-bottle quotienting) and taking union over the resulting sets (and $V$), we obtain a new closed set that intersects $X$ precisely in $D$. Try drawing the picture.

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Thank you! What is $X^0$? –  goobie Feb 5 '13 at 19:11
    
@goobie: Interior. I thought this was standard notation, but now I am not sure. –  Marek Feb 5 '13 at 19:20
    
Thank you, yes it is standard notation. But I wasn't sure because the symbol you are using seems to be the zero whereas the symbol I keep seeing used more commonly is $\mathring{X}$. –  goobie Feb 5 '13 at 19:23
    
By the way: In the first paragraph, did you mean to write "...and $q(C)$ is closed precisely when $D$ is..."? –  goobie Feb 5 '13 at 19:25
    
I am stuck on your last paragraph. The set $W$ lies outside $X$. On the other hand, $D$ lies inside $X$ but what I cannot work out is why we can "move" $W$ so that it covers $D$? They don't have the same shape. What am I doing wrong? –  goobie Feb 5 '13 at 19:30
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Define the following closed subsets of the square $M$: The four corner points $K$, the top boundary $T \cong [0,1]$, the bottom boundary $B\cong [0,1]$, the left boundary $L\cong [0,1]$ and the right boundary $R\cong [0,1]$. Note that reflection at either the line $y=\frac{1}{2}$ or $x = \frac{1}{2}$ is a homeomorphism. If $C$ is closed in $M$ then $D$ is either equal to $C$ or it is equal to $C$ union a reflected copy of $C$ intersected with one of the closed sets $T,B,L,R,K$. Hence $D$ is a union of closed sets and hence closed.

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