Lattice for topology

Question

I have been reading Clint Enns (https://mspace.lib.umanitoba.ca/bitstream/handle/1993/21613/Enns_Pure_embeddings.pdf?sequence=1&isAllowed=y) and found Theorem 3.3 in page 96 that state "The set of topologies that make $M$ a topological module forms a complete lattice with respect to $\subseteq$". In the proof of this theorem: Let $\mathcal{T}=\{T_{i}\}$ be the set of all topologies that make $M$ a topological modules and $\mathcal{A}\subseteq \mathcal{T}$. The join of $\mathcal{A}$ is $\tau$ topology generated by $\cup \mathcal{A}$ as a subbase. One of the continuity, it proves that $-\colon M\to M$, $m\mapsto -m$, is continuous function with respect to $\tau$. If $V$ is open set in $M$, we have to prove that $(-)^{-1}(V)$ is open in M. For such a $V$, $V\in T_{i}$ for some $i\in I$. SInce $-$ is continuous with respect to $T_{i}$, $(-)^{-1}(V)\in T_{i}\subseteq\tau$. Therefore, $-$ is continuous with respect to $\tau$. enter image description here When I read this proof, I really didn't get it, why if we have $V$ is open set in $M$ respect to $\tau$ topology generated by $\cup \mathcal{A}$ as a subbase, then we have $V\in T_{i}$ for some $i\in I$. I mean, if we have $\tau$ topology generated by $\cup \mathcal{A}$ as a subbase, we have \begin{align*} \mathcal{S}=\cup\{A_{i}\mid A_{i}\in\mathcal{A}\}, \text{ is subbase}\\ \mathcal{B}=\cap\mathcal{F}, \mathcal{F}\subseteq\mathcal{S}, \mathcal{F}\text{ is finite subset of }\mathcal{S}\\ \tau=\cup\{B\mid B\in \mathcal{B}\} \end{align*} So, when we have $V\in\tau$, we can write $V=\cup_{B\in\mathcal{B}}B$ where $B=\cap_{i=1}^{n}S_{i}$, $S_{i}\in \mathcal{S}$. How we have, $V\in T_{i}$ for some $i\in I$? Thank you for any help.

Answer 1

Basic lemma (implicit in the proof):

If $f: (X, \mathcal{T}_X \to (Y, \mathcal{T}_Y)$ is a function between spaces and $\mathcal{S}$ is a subbase for $\mathcal{T}_Y$ then $f$ is continuous iff $f^{-1}[S]$ is open in $X$ for all $S \in \mathcal{T}_Y$.

Proof: the left to right implication is obvious as all inverse images of open sets are open, in particular the elements of $\mathcal{S}$ (which are open in $Y$ by definition). For the right to left implication: let $O$ be in $\mathcal{T}_Y$. By standard facts about subbases, we can write $O$ as a union of $O_i, i \in I$ ($I$ some index set) where each $O_i$ is a finite intersection $\bigcap \mathcal{S}'_i$ for some finite $\mathcal{S}'_i \subseteq \mathcal{S}$. By assumption all $f^{-1}[S]$ are open in $X$ for all $S \in \mathcal{S}$ and by standard set theory

$$f^{-1}[O]= f^{-1}[\bigcup_{i \in I} O_i] = \bigcup_{i \in I} f^{-1}[O_i] = \bigcup_{i \in I} f^{-1}[ \bigcap \mathcal{S}'_i ] = \bigcup_{i \in I} \bigcap \{ f^{-1}[S]: S \in \mathcal S'_i\} $$

which is open as $\mathcal{T}_X$ is closed under finite intersections and all unions. So $f$ is continuous.

Your continuity proof is just this lemma applied to $f=-$ and $\mathcal{S} = \bigcup \mathcal{A}$. All sets in $\bigcup \mathcal{A}$ are in some module-topology $\mathcal{T}_i$ from the join-forming family $\mathcal{A}$ and $f$ is continuous wrt all such topologies. So inverse images of subbasic sets are open (even subbasic open again, namely in the same $\mathcal{T}_i$ so in $\bigcup \mathcal{A}$).