I'm asking for feedback on my current work.
Preamble: It is known that if $(f_i)_{i\in I}$ a family of functions $f_i:X\to Y_i$, where $(Y_i, \mathcal{T}_i)$ is a topological space, then the topology $\mathcal{T}$ induced by this family is the coarsest topology with which every $f_i:X\to Y_i$ is continuous.
Here's one
Claim: Here's one source for the claim, albeit it seems to leave the uniqueness out: Universal Property
Let $X$ be a set and $(f_i)_{i\in I}$ a family of functions $f_i:X\to Y_i$ with $(Y_i, \mathcal{T}_i)$ being a topological space for all $i \in I$. Then the topology $\mathcal{T}$ induced by $(f_i)_{i\in I}$ for $X$ is the only topology for $X$ with the following property: Let $(Z, \mathcal{T}')$ be a topological space and $g:Z\to X$ a function. Then $g$ is continuous iff $f_i\circ g$ is continuous for all $i \in I$.
Suppose that the equivalence of the continuities has already been proven. Therefore, what remains is to show that $\mathcal{T}$ is the only topology with the continuity property.
My work: As I've understood the setup, I have to show that if $\mathcal{T}_2$ is any other topology with satisfies the said continuity property, then necessarily $\mathcal{T}_2 = \mathcal{T}$. So let $\mathcal{T}_2$ be such a topology. Since the said property must hold for all $(Z, \mathcal{T}')$ and $g$, we may choose $g = \mathrm{id}, Z = X, \mathcal{T}' = \mathcal{T}$.
Then, if $g$ is continuous, it follows that all neighborhoods of $\mathcal{T}_2$ are open in $\mathcal{T}'$, i.e. $\mathcal{T}_2 \subset \mathcal{T}'$. Since $g$ is continuous then also $f_i \circ g = f_i$ is continuous as well. But since now $f_i$ is continuous w.r.t. $\mathcal{T}_2$ and $\mathcal{T}$ was the coarsest topology with which $f_i$ is continuous, it follows that $\mathcal{T}\subset \mathcal{T}_2$. Hence $\mathcal{T} = \mathcal{T}_2$.
Thoughts: Is the argument above enough? I don't think that I've ever proven a claim that some object $X$ is the only one which satisfies an equivalence predicate, so I'm not sure is there more work to do. Namely, I worked through the one way of the equivalence, assuming that $g = \mathrm{id}$ is continuous. Should I also show that $\mathcal{T}_2 = \mathcal{T}$ when it is assumed that $f_i \circ g$ is continuous?
