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A continuous map and surjective f: X $\rightarrow$ Y $\Longrightarrow$ $\tau_Y =$ $\tau_Q$, where $\tau_Y$ is an arbitrary topology and $\tau_Q$ is a quotient topology?

I tried proof this, but I don't sure.

My attempt:

$\tau_Y \subseteq$ $\tau_Q$

Let be $V \in \tau_Y$, $f$ is continuous map, so $f^{-1} (V)$ $\in$ $\tau_X$ and $\tau_Q =$ { $V \subseteq Y$ ; $f^{-1} (V)$ $\in$ $\tau_X$ and f is surjective}, therefore $V \in \tau_Q$.

$\tau_Q \subseteq$ $\tau_Y$

Let be $V \subseteq Y$ ; $f^{-1} (V)$ $\in$ $\tau_X$. As the $f$ is a continuous map, $V \in \tau_Y$.

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  • $\begingroup$ As shown by answers below, the answer is no. It is true when $f$ is assumed to be an open or closed map as well. But not in general. $\endgroup$ – Henno Brandsma Jan 15 '16 at 12:54
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That $f: X \rightarrow Y$ is continuous means: if $V \subseteq Y$ is open, then $f^{-1}[V]$ is open as well. And $Y$ has the quotient topology means exactly that the reverse implication also holds: namely that if $f^{-1}[V]$ is open in $X$, then $V$ is open in $Y$.

And being surjective is not enough to garantuee this. E.g. if $f$ were an open map, then we conclude from $f^{-1}[V]$ open in $X$ that $f[f^{-1}[V]]$ is open in $Y$ (by open-ness of $f$!) and $f[f^{-1}[V]] = V$ when $f$ is surjective. So an open, continuous, surjective map has the property that it is quotient.

And as we can equivalently formulate everything in terms of closed sets and their inverse images, we can also say that a closed, continuous, surjective map is quotient.

Counterexamples:

$f(x) = x$ from $X = \mathbb{R}$ in the discrete topology, and $Y = \mathbb{R}$ in the usual topology. Then $f^{-1}[\{0\}]$ is open in $X$ (everything is open in $X$) but $\{0\}$ is not open in $Y$.

$f(x) = (\cos(x), \sin(x))$ from $X = [0,2\pi)$ to $Y = S^1$ (the unit circle in the plane), both in their usual topology. Then $f$ is not a quotient map, as otherwise $f$ would be a homeomorphism (a 1-1 and onto quotient map always is) and this cannot be, by the usual arguments.

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  • $\begingroup$ I understood, thank you! $\endgroup$ – George Jan 16 '16 at 22:28
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Your proof for $\tau_Q\subseteq\tau_Y$ is wrong. Since $f$ is continuous, you know that if $V\in \tau_Y$, then $f^{-1}(V)\in\tau_X$. But what you're trying to use is the converse of this, and there's no reason for the converse to be true.

In fact, the result is not true in general. To find a counterexample, I suggest looking at the case when $f$ is a bijection.

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