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Let $(X,\tau_x),(Y,\tau_Y)$ be topological spaces and let $f:X \to Y$ be a continuous and open function. Prove that $f(\tau_X)=\tau_Y$

This is an excercise of the problems set of my course on Introductory Topology notes. I think what I've been asked to prove is not true since if I chose $X,Y$ both equiped with the discrete topology and $Y$ having more than one element, then any constant function $f$ $(f(x)=c \in Y$ for all $x \in X)$ is continuous and open but $f(\tau_X)=\{ \emptyset, \{c\} \}\not=\tau_Y$
Is this right? Perhaps I'm missunderstanding it. Any clarification or suggestion is welcome. Thanks

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Your counterexample is good. You could also consider the fact that $Y\notin f(\tau_X)$ unless $f$ is surjective; for instance, the inclusion of a proper open set is continuous and open.

Suppose that $f$ is surjective, besides being open and continuous. It's clear that $f(\tau_X)\subseteq \tau_Y$ (by the fact $f$ is open). If $V$ is open in $Y$, then $V=f(f^{-1}(V))$ (by surjectivity of $f$), so $V\in f(\tau_X)$ (by the fact $f$ is continuous).

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    Many texts assume that open maps are onto. That also ensures they are quotient maps. – 2017-02-18
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    @HennoBrandsma That's why I added the proof for the case when the map is surjective. – 2017-02-18
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    I figured as much, yes. – 2017-02-18
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Right. In fact, a continuous function is open if and only if $f(\tau_X) \subseteq \tau_Y$, but the converse inclusion needn't hold, in general.

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    I see, thanks for the clarification. Actually the excercise has a second part, which asks: Is it always true that $f^{-1}(\tau_Y)=\tau_X$? I think the problem should have been stated the other way round. Probably it was intended to be : "prove $f^{-1}(\tau_Y)=\tau_X$, and is it always true that $f(\tau_X)=\tau_Y$" – 2017-02-17