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We know that choice of topology on two set is induces a choice of which maps are continuous between two sets. Suppose we knew all the functions between two topological spaces of unknown topology which are continuous, would we be able to deduce the topology on the sets back?

I think so the answer is yes, but I am a bit bogged down on what exactly giving all the continuous function between two topological spaces entails.

tryst with freedom
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    Let's say the functions are $X\to Y$. There are notable cases where you know the continuous functions $X\to Y$ and the topology on $Y$, but you hardly know anything about the topology on $X$. Namely: if $Y$ is empty or a point you can only know if $X$ is empty; if $Y$ is a discrete space with at least two points, $X\ne\varnothing$ and the continuous functions are the constants you only know that $X$ is connected. – Sassatelli Giulio May 15 '22 at 07:37
  • Assume $X$ and $Y$ are topological spaces with at least two elements (otherwise there is only one topology). If every map $f\colon(X,\mathcal{O}_X)\rightarrow(Y,\mathcal{O}_Y)$ is continuous, then we could have $\mathcal{O}_X=\mathcal{P}(X),\mathcal{O}_Y=\mathcal{P}(Y)$ or $\mathcal{O}_X=\mathcal{P}(X),\mathcal{O}_Y={\emptyset,Y}$ or $\mathcal{O}_X={\emptyset,X},\mathcal{O}_Y={\emptyset,Y}$. In this case, we can't deduce the topology and need further conditions. – Samuel Adrian Antz May 15 '22 at 07:43
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    Knowing the topology on $Y$, you may deduce some "minimal" topology on $X$ that is consistent with all those functions being continuous; sometimes it would be the original topology of $X$, and sometimes it wouldn't. Not knowing the topology on $Y$, you can't do even that. – Ivan Neretin May 15 '22 at 09:28
  • Welp three good comments, I would have accepted them if they were answer – tryst with freedom May 16 '22 at 09:01
  • A possible reformulation, in light of the above comments, could be as follows: Let $X$ be a topological space with unknown topology $\mathcal{T}(X)$. Call a pair $\mathscr{P}=(\mathscr{L},\mathscr{R})$ a probe for $X$ if $\mathscr{L}$ is a subcollection of the slice category $Top/X$ and $\mathscr{R}$ is a subcollection of the slice category $X/Top$. What are (minimal) sufficient conditions on probes that determine $\mathcal{T}(X)$ uniquely? – Alp Uzman May 18 '22 at 18:07
  • The right probe $\mathscr{R}$ gives immediately a lower bound, using the left probe seems to be more subtle. – Alp Uzman May 18 '22 at 18:08
  • See also https://math.stackexchange.com/q/756115/169085 – Alp Uzman May 18 '22 at 18:12
  • One can also consider a higher analog of the probe formalism, e.g. say we're given two topological spaces $X$ and $Y$ with unknown topologies. Call a triple $\mathscr{P}=(\mathscr{L},\mathscr{M},\mathscr{R})$ a probe for $X\to Y$ if $\mathscr{M}\subseteq C^0(X,Y)$, any element of $\mathscr{L}$ is in $HomTop/f$ for some $f\in\mathscr{M}$, and any element of $\mathscr{R}$ is in $f/HomTop$ for some $f\in\mathscr{M}$. – Alp Uzman May 18 '22 at 18:20
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    Hi!! @AlpUzman some part of what you said are beyond my correct topology knowledge. Thank you so much for your interest in my question though :) – tryst with freedom May 18 '22 at 19:34

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