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It's well-known that if a continuous function taking values in a Hausdorff space is uniquely determined by its specification on a dense subset of the domain. Now, I contemplate on the necessity of Hausdorff-ness in this result:

It's clear that the result no longer holds if the codomain is just $T_0$, as is demonstrated by the function $\mathbb R$ to the Sierpiński space $\{0, 1\}$ (with $1$ being the Sierpiński point) given by $x\mapsto 0$ if $x\ne 0$ and $x\mapsto 1$ if $x = 0$. (Thus this and the constant $0$ function are both continuous despite agreeing on the dense $\mathbb R\setminus \{0\}$.)

Now, I am trying to come up with an similar example where the codomain is $T_1$ (and not Hausdorff ofc), but haven't been able to conjure anything up yet. You know anything? The only examples of $T_1$ spaces fmailiar to me are the co-finite/-countable spaces.

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    $\begingroup$ Hello. Are you aware that $X$ being Hausdorff is equivalent to $\{y \in Y: f(y) = g(y)\}$ being closed for all topological spaces $Y$ and functions $f, g: Y \to X$? This is usually proved via the also equivalent condition "the diagonal $\{(x, x) : x \in X\}$ is closed in $X^2$". I'm struggling a bit to find a good link to a proof of that at the moment, but it answers your question very strongly - if $X$ is not Hausdorff, you can find $f$ and $g$ which agree on a non-closed set. In particular, that set is dense in its closure but $f$ and $g$ do not agree on the closure! $\endgroup$ Feb 22 at 15:00
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    $\begingroup$ I've found some links now. You can piece it together from here and here. It's also mentioned here. When you piece it together, what you end up with is "if $X$ is not Hausdorff, then the two projection maps $\pi_1, \pi_2 : X \times X \to X$ agree on the diagonal $\Delta \subseteq X \times X$, but not on its closure". I'm not sure if this is close enough to count as a duplicate or not. $\endgroup$ Feb 22 at 15:14
  • $\begingroup$ @IzaakvanDongen Thanks tons for your responses! But I think I have found a simple example: Any bijection on $\mathbb R$ is continuous with the codomain being taken under cofinite topology. Now, just consider the identity function with two points swapped. What do you think? $\endgroup$
    – Atom
    Feb 22 at 15:55
  • $\begingroup$ @Atom Looks good to me, with the dense subset taken to be $\Bbb R\setminus\{a,b\}$ where $a,b$ are the swapped points $\endgroup$
    – FShrike
    Feb 22 at 16:10
  • $\begingroup$ @FShrike Great, I will post as an answer then $\endgroup$
    – Atom
    Feb 22 at 16:38

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Any bijection on $\mathbb R$ is continuous with the codomain taken under cofinite topology. Now, just consider the identity function with two distinct points, $a$ and $b$, swapped. Then this and the identity function are continuous functions agreeing on the dense subset $\mathbb R\setminus\{a, b\}$ and yet distinct.

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