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I've seen in various books that, in order to prove that a function $f:X\rightarrow Y$ is continuous, it is sufficient to show that the pre-images $f^{-1}(V)$ of all sets $V$ of a certain basis of the topology in $Y$ are open (in $X$). This is clear to me and easy enough to prove.

I would say that the reciprocal is also true. i.e. $f$ is continuous iff $\forall \, V\in \beta_Y,\,f^{-1}(V)\in T_X$ (with $\beta_Y$ a basis of $Y$ and $T_X$ the topology on $X$). This comes simply from the fact that all the elements of the basis $\beta_Y$ are elements of the topology $T_Y$.

Now, I've also seen that in order to prove that a function $f:X\rightarrow Y$ is continuous, it is sufficient to show that the pre-images $f^{-1}(V)$ of all sets $V$ of a subbasis of the topology $T_Y$ are open (in $X$). This is also clear.

My question: is the reciprocal of this last statement true, like in the basis case? i.e. is it true that $f$ is continuous iff $\forall \, V\in \gamma_Y,\,f^{-1}(V)\in T_X$ (with $\gamma_Y$ a subbasis of $Y$ and $T_X$ the topology on $X$)? My intuition tells me that this is false: the elements of the subbasis $\gamma_Y$ are not elements of the topology $T_Y$, so I don't see any reason for having $\forall \, V\in \gamma_Y,\,f^{-1}(V)\in T_X$ just because $f$ is continuous.

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    Elements in a subbasis are open, since finite intersections of them form a basis. – 2017-01-16
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    @AndréS. Finite intersections of open sets are open themselves, but just because a certain open set $A$ is the finite intersection of other sets, does that mean that these other sets are open? I don't think so: just take the set $X=\{a,b,c\}$ with topology $\{\phi ,X\}$. The intersection of the sets $\{a\}$ and $\{b\}$ is $\phi$ and hence open, but $\{a\}$ and $\{b\}$ are not open. – 2017-01-16
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    If $S$ is an element of a sub basis for the topology $T$, then $S$ is itself a finite intersection of elements from the subbasis (e.g. $S\cap S$), so it is open. – 2017-01-16
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    This is the whole point of subbases: The topology generated by a subbasis $\mathcal{S}$ is the smallest topology with the property that every set in $\mathcal{S}$ is open. – 2017-01-16
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    @AlexKruckman Ok, I think I got it. In this specific case, then, the last "iff" is also true, then. Right? – 2017-01-16
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    Yes, that's right. – 2017-01-16

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