# Why is this a topology on $Y$?

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 $Y$?

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I can't prove that $\tau _q$ is closed under finite intersection. – user73564 May 26 '13 at 16:42
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
Yes, so what does it mean for $f$ to be a quotient map? – Ted Shifrin May 26 '13 at 17:06

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.
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