Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a continuous map from $X$ onto $Y$, and $f$ is a quotient map. Why is the family $\tau_q $ of all $f(U)$, where $U$ is open in $X$, a topology on $Y$?

share|improve this question
2  
What have you tried so far? –  Daniel Rust May 26 '13 at 16:41
2  
I can't prove that $\tau _q$ is closed under finite intersection. –  user73564 May 26 '13 at 16:42
2  
I am confused by the hypotheses. You say $f$ must be a cts map, which implies that $Y$ came with a topology. But this can't possibly be used anywhere, since we are only being asked to check that another collection of subsets of $Y$ forms a topology. –  user29743 May 26 '13 at 16:44
    
All sets $O$ that are open in the original topology on $Y$ (which must be assumed to exist or continuity of $f$ does not make any sense) are open under this definition, as for onto maps $O = f[f^{-1}[O]]$. Don't yet see how to use this yet. –  Henno Brandsma May 26 '13 at 16:55
    
Does change something if we suppose that $f$ is a quotient map? (I edited my original post) –  user73564 May 26 '13 at 17:04
show 5 more comments

1 Answer

It is not, in fact, a topology. For a simple counterexample, let $X=\{1,2,3,4\}$ with the topology $\{ \{1,2\}, \{3,4\}, X, \emptyset\}$. Let $Y=\{1,2,3\}$ with the indiscrete topology. Define $f$ as $f(1)=1, f(2)=f(3)=2, f(4)=3$. Now, $$ \tau_q = \{ Y, \emptyset, \{1,2\}, \{2,3\}\}$$ which is not closed under intersection.

Edit: $f$ is also a quotient map.

share|improve this answer
    
We can suppose in addition that $f$ is a quotient map, just in case that it changes something... thank you –  user73564 May 26 '13 at 17:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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