Give an explicit characterization of open sets in the weak topology on $X$ induced by a single map $f : X→ Y$ Let $X$ be an arbitrary set. There is a natural way in which a family of maps
$f_j : X → Y_j$ (for $j ∈ J$) from $X$ into topological spaces $Y_j$ induces a topology on $X$. Namely, the weak topology
induced by the family {$f_j : j ∈ J$} is the coarsest topology on $X$ with respect to which all the maps $f_j$ are
continuous.
$(a)$ Verify that the product topology on the product $\prod_{i∈I} X_i$ of a family of topological spaces is the weak topology induced by the projections $π_k :\prod_{i∈I} X_i → X_k$ (for $k ∈ I$).
$(b)$ Verify that the subspace topology on a subset $A ⊆ X$ of a topological space X is the weak topology induced
by the inclusion map $ι_A : A → X$ (that is the single-element family {$ι_A$}).
$(c)$ Give an explicit characterization of open sets in the weak topology on $X$ induced by a single map $f : X→Y$ into a topological space $Y$.

I proved parts $(a)$, and $(b)$, but I'm stuck on part $(c)$. Any help please?

 A: Not a characterisation but an observation: if $f: X \to (Y, \mathcal{T}_Y)$ is given, the weak topology wrt $f$ on $X$ is exactly
$$\{f^{-1}[O]\mid O \in \mathcal{T}_Y\}$$
as any such set must be in the topology (no choice) to get $f$ continuous, and the collection is already a topology because $f^{-1}$ behaves nicely wrt set operations, e.g.
$$f^{-1}[\bigcup_i O_i] = \bigcup_i f^{-1}[O_i]$$ and similarly for intersections.
As for any weak topology we have the continity characterisation,
$g: (Z, \mathcal{T}_Z) \to X$ is continuous wrt to the weak topology on $X$ induced by $f$ iff $f \circ g:(Z, \mathcal{T}_Z) \to (Y, \mathcal{T}_Y)$ is continuous. See this post of mine on initial topologies (another name for weak topologies).
But if $(c)$ is asking the more general question that

given a space $(X,\mathcal{T}_X)$ can we find a necessary and sufficient condition on this space so that there exists some space $Y$ and some function $f: X \to Y$ such that the given topology on $X$ is exactly the weak topology wrt $f$?

then I don't know if such a characterisation exists. Maybe all topologies can be obtained that way...
But if $f$ is given, the first paragraph can be seen as a sort of characterisation of the weak topology: every open set is exactly the inverse image under $f$ of a $Y$-open set.
Probably that is the intended answer here.
