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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?

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  • $\begingroup$ It seems that your main goal is checking this specific proof - so perhaps (solution-verification) would be a suitable tag here. (See the tag-info for more details about the usage of this tag.) I think that also (universal-property) could fit. It is also worth pointing out that in the title you mention the subspace topology, while the whole post seems to be about the initial topology. $\endgroup$ Mar 17 at 13:06
  • $\begingroup$ I think that you could reorder the statement, and put $\mathcal T$ at the end of the claim, so there would be no confusion on how to prove it. Just like it was mentioned here. As Martin Sleziak has said, you are working with the initial topology, <- here Henno Brandsma explains very concisely what this is; and in this same answer Henno proved the unicity. $\endgroup$ Mar 18 at 12:11

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