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Can somebody please explain me the following example given in Munkres, quotient topology?

Let $A={a,b,c}$ and $p$ be map from real line to the set $A$ defined by

$p(x)=a$ if $x>0$, $p(x)=b$ if $x<0$, $p(x)=c$ if $x=0$.

Find the quotient topology on $A$ induced by $p$.

I understand the definition of quotient topology,but couldn't get an idea about how to form the topology given in this example.

How can we relate this example to quotient space?

Here the subsets of $A$ are $\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\},\{\}$.

by definition of $p$, inverse image of $\{a\}$ is the open interval $(0,\infty)$, inverse image of $\{b\}$ is the open interval $(-\infty,0)$ and inverse image of $\{a,b\}$ is $\Bbb{R}\setminus\{0\}$.

for other subsets we cannot find any open set in $\Bbb{R}$ such that the inverse maps to that open set.

Is this sufficient for us to say that the topology induced by $p$ is a topology on $A$?

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Do you seriously expect people to take the time to go and look up that example to help you if you're not even willing to invest the time to quote it? –  joriki Nov 10 '12 at 11:48
    
it would be very helpful if you could provide a more helpful title, and the actual example, because I don't have the book. –  akkkk Nov 10 '12 at 11:48
    
Can you cite here briefly that example? –  Berci Nov 10 '12 at 11:48
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While Munkers is quite a popular book please don't rely on the reader to have it at hand. Furthermore please don't force the reader to get up from their chair, find the book, flip through the pages, and then reach the final page just to find the example. Just write the example here. If you put effort into the questions, people will put effort into their answers. –  Asaf Karagila Nov 10 '12 at 11:49

1 Answer 1

up vote 1 down vote accepted

You only have to check that $p^{-1}(b)= ]- \infty,0[$ is open, $p^{-1}(a)=]0,+ \infty[$ is open, $p^{-1}(c)= \{0 \}$ is not open, $p^{-1}(\{a,b\})= \mathbb{R}^*$ is open, $p^{-1}(\{a,c\})=[0,+ \infty[$ is not open, etc.

Indeed, $U \subset A$ is open iff $p^{-1}(U)$ is open in $\mathbb{R}$.

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