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Say we have a space $X$ and a collection of functions $\{f_i:X \to Y\}$, where $Y$ is endowed with some topology $\mathcal{T}_Y$, and $\mathcal{T}$ is the smallest topology making all $f_i$ continuous. Then $\mathcal{T}$ is generated by $S := \{f_i^{-1}(V) : V \in \mathcal{T}_Y \}$.

Then, if we have another map $g:Y\to X$, and we wanted to show that $g$ is continuous, i.e. that for every open in $X$, its preimage is open in $Y$. Would it be sufficient to only show this for the opens in $S$, i.e. that for all $U \in S$, $g^{-1}(U) \in \mathcal{T}_Y$?

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Yes. more generally, Let $(X,\tau_1)$ and $(Y,\tau_2)$ be topological spaces. if $\tau_1$ is generated by a base $\cal B$ then $$f:Y\to X$$ is continues iff $f^{-1}(U)\in \tau_2$ for all $U\in \cal B$.

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  • $\begingroup$ yes!edited, thanks $\endgroup$
    – sha
    Commented Jan 19, 2015 at 22:59

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