Let $q:X\to Y$ be a surjective map, where $(X,\tau_X)$ is a topological space. The quotient topology $\tau_q$ on $Y$ is given as $U\in \tau_q$ iff $q^{-1}(U)\in \tau_X$.

Suppose that there is another topology $\tau_d$ on $Y$ such that for any topological space $Z$ is holds that $f:Y\to Z$ is continuous iff $(f\circ q):X\to Z$ is continuous. I should prove that then $\tau_d = \tau_q$.

My idea is to consider $f_1 = \operatorname{Id}_Y:Y_d\to Y_q$ and $f_2 = \operatorname{Id}_Y:Y_q\to Y_d$ to use the characteristic property. Then I have that since $q:X\to Y_q$ is a quotient map, it is continuous hence $f_2$ is continuous. Here $Y_q = (Y,\tau_q) $ and $Y_d = (Y,\tau_d)$.

But I don't have an idea how to prove that $f_1$ is continuous since I don't know that $q:X\to Y_d$ is continuous. To prove the last fact I consider $f_3 = \operatorname{Id}_Y:Y_d\to Y_d$ which is continuous so is $q = (f_3\circ q):X\to Y_d$. Hence, $q:X\to Y_d$ is continuous so is $f_1$.

Could you please help me to realize if my proof is correct?

  • 1
    $\begingroup$ Presumably when your write ${\tau}_Y$ in paragraph 2 you mean ${\tau}_d$. $\endgroup$ – Thomas Andrews Oct 24 '11 at 15:34
  • $\begingroup$ @ThomasAndrews: you're right, thanks. Just tried to improve the notation and forgot about that place. $\endgroup$ – Ilya Oct 24 '11 at 15:43

To avoid any confusion use distinct symbols for the mapping $q$ depending on whether the image is $Y_q$ or $Y_d$. So denote $q'=f_2\circ q:X\to Y_d$.

To show that $f_1$ is continuous use the property of $\tau_d$. We know that the quotient map $q=f_1\circ q':X\to Y_q$ is continuous. Hence $f_1$ is continuous by the backward direction of the property of $\tau_d$.

To show that $f_2$ is continuous first use the forward direction of the property of $\tau_d$ to show that $q'=f_3\circ q'$ is continuous (as you have done). Then noting that $q'=f_2\circ q$ use the characteristic property of the quotient space $Y_q$ to conclude that $f_2$ is continuous.

Finally the continuity of $f_1$ and $f_2$ implies that $\tau_d=\tau_q$.


(Your proof is essentially correct, but there might be a clearer way to present it.)

I'd say the clearest way is to separate this into two problems.

(1) There can be at most one topology on $Y$, $\tau$, such that for all spaces $Z$, a function $g:Y\rightarrow Z$ is continuous if and only if $g\circ f:X\rightarrow Z$ is continuous.

This essentially proves that ${\tau}_d$ in your problem is unique if it exists.

(2) ${\tau}_q$ has the property described in (1).

Proof of (1): If ${\tau}$ has this property, then, since $id_\tau:(Y,\tau)\rightarrow(Y,\tau)$ is continuous, $f_\tau:X\rightarrow(Y,\tau)$ is continuous. Now, given $\tau$ and $\tau'$ topologies on $Y$ with this property, we need to show that ${id}_1:(Y,\tau)\rightarrow (Y,\tau')$ and ${id}_2:(Y,\tau')\rightarrow (Y,\tau)$ are continuous. But a function from $(Y,\tau)$ is continuous if any only if the composition with $f_\tau$ is continuous, and we know that $f_\tau \circ {id}_1 = f_{\tau'}$, which we know is continuous. Similarly, ${id}_2$ is continuous, so $\tau=\tau'$.

Proof of (2): If $g:(Y,{\tau}_q) \rightarrow Z$ is continuous, then $V$ is open in $Z$ implies that $g^{-1}(V)\in {\tau}_q$ which implies that $f^{-1}(g^{-1})(V)$ is open in $X$. But that means that $g\circ f$ is continuous.

On the other hand, given $g:Y\rightarrow Z$, if $g \circ f$ is continuous, then for any open $V$ in $Z$, $(g\circ f)^{-1}(V) = f^{-1}(g^{-1}(V))$ is open in $X$. But, by the definition of ${\tau}_q$, that means that $g^{-1}(V)\in {\tau}_q$, so $g:(Y,\tau_q)\rightarrow Z$ is continuous.

So $\tau_q$ has the property from (1).

Note that this proof is a stronger result than the original problem, because the original problem essentially says the only candidate topology for (1) is $\tau_q$. It might be that no topology satisfies (1), in which case, your original problem would still be true.


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