A question related to the smallest topology on a set.

Question

Let $X$ be a non-empty set and $((Y_\lambda,\tau_\lambda))_{\lambda\in\Lambda}$ a family of topological spaces. Suppose $(f_\lambda:X\rightarrow Y_\lambda)_{\lambda\in\Lambda}$ is a family of functions. Let $\mathcal{S}=\{f_{\lambda}^{-1}(G)|\lambda\in\Lambda\ and\ G\in\tau_\lambda\}$.

Now it is required to show that the topology $\tau$ on $X$, for which $\mathcal{S}$ is a sub basis, is the smallest topology on $X$ such that $f_\lambda$ is continuous $\forall\lambda\in\Lambda$.

Can someone help me answer this question please? A hint is appreciated. Thank you.

Answer 1

This is known as the initial topology.

If $(X,\tau)$ is a topological space so that all $f_\lambda \to Y_\lambda$ is continuous, then $f^{-1}(U)$ is open for all open $U \subseteq Y_\lambda$.

One can prove the statement by showing that the topology generated by a sub-basis is the smallest one containing all of the open sets in $S$ (since one can describe the generated topology as the intersection of all topologies containing $S$), which is a more general fact.

There are more details written here or here.