Meaning of "the weakest topology such that <blank> is continuous"

If $τ$ is the weakest topology on $X$ such that $f:X→Y$ is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets under $Y$? This follows directly from the definition of a "continuous" function. Is this always the coarsest topology?

Yes. For $f$ to be continuous, you need the topology on $X$ to contain all preimages of open sets through $f$. The topology induced by a family $\mathcal T$ of functions is generated by $$ \{f^{-1}(E):\ f\in\mathcal T,\ E\subset Y\ \text{ open }\}. $$ Some reasons why one cares about these topologies are

• They often appear naturally, as when considering duals and preduals of normed spaces;

• In several cases the topology is coarse enough that some interesting sets become compact (for instance the unit ball in a Banach space, see the Banach-Alaoglu Theorem).

"weakest"

Fix a space X, and let Top(X) be the set of topologies on X. That is, an element T of Top(X) can be thought of as a subset of the power set P(X) (satisfying axioms). Now, Top(X) naturally forms a poset under $\subseteq$: that is, if S and T are elements of Top(X), you can think of S as "smaller than" T if $S\subseteq T$, i.e. all S-open sets are also T-open (but not necessarily conversely). The indiscrete topology is the "smallest" element of this poset, and the discrete topology is the "largest" element.

The words "coarse" and "fine" are synonyms in this context for "small" and "large" respectively. That is, the indiscrete topology is the coarsest possible topology (it has very few open sets; it smooshes all points of X together into one big open set, so that you can't tell the points or subsets apart with open sets), and the discrete topology is the finest possible topology (it has lots of open sets; you can tell lots of things apart). If X is e.g. $\mathbb{C}^n$, then you could also define the Euclidean topology, or the (weaker) Zariski topology.

Note the following: if S and T are both elements of Top(X), then $S\cap T$ is too. (Prove it!)

The "weakest" topology subject to some conditions, then, is the smallest possible topology that satisfies those conditions. Put another way, it's the intersection of all the topologies that satisfy those conditions. Put yet another way, if you are happier with algebra than topology, you might like to think of this as the topology generated by some open sets.

$f: X\to Y$ is continuous if, for every open U in Y, the set $f^{-1}(U)$ is open in X. So the weakest topology on $X$ such that $f:X\to Y$ is continuous is the smallest one that contains $f^{-1}(U)$ for every open U in Y. That's the intersection of all those T in Top(X) such that $f^{-1}(U)\in T$ for all open U in Y. And so on.