Vtyjkyuk

$let op operatorname$There are several characterizations of continuity of a map $f: X to Y$ between topological spaces of the form $f^•[opop A] diamond opop f^•[A]$ for every $A ⊆ X$ or $Y$ where $f^• ∈ f, f^-1, f^∀$ denotes image, preimage, and dual image ($f^∀[A] := Y setminus f[X setminus A]$), $opop ∈ opcl, opint$ denotes the closure or interior operator, and $diamond ∈ ⊆, ⊇$ denotes the inclusion in either direction. This gives $3 × 2 × 2 = 12$ combinations. By using some symmetries and dualities, it is easy to show that some conditions are equivalent.

Namely, the following conditions correspond to continuity, i.e. $f^-1[U]$ is open for every open $U ⊆ Y$, or equivalently $f^-1[F]$ is closed for every closed $F ⊆ Y$:

• $f[opcl(A)] ⊆ opcl(f[A])$,


• $f^∀[opint(A)] ⊇ opint(f^∀[A])$,


• $f^-1[opcl(B)] ⊇ opcl(f^-1[B])$,


• $f^-1[opint(B)] ⊆ opint(f^-1[B])$.

The following conditions correspond to closedness, i.e. $f[F]$ is closed for every closed $F ⊆ X$, or equivalently $f^∀[U]$ is open for every open $U ⊆ X$:

• $f[opcl(A)] ⊇ opcl(f[A])$,


• $f^∀[opint(A)] ⊆ opint(f^∀[A])$.

The following conditions correspond to openness, i.e. $f[U]$ is open for every open $U ⊆ X$, or equivalently $f^∀[F]$ is closed for every closed $F ⊆ X$:

• $f[opint(A)] ⊆ opint(f[A])$,


• $f^∀[opcl(A)] ⊇ opcl(f^∀[A])$,


• $f^-1[opint(B)] ⊇ opint(f^-1[B])$,


• $f^-1[opcl(B)] ⊆ opcl(f^-1[B])$.

The remaining two conditions are also equivalent and correspond to what I call “co-continuity” here:

• $f[opint(A)] ⊇ opint(f[A])$,


• $f^∀[opcl(A)] ⊆ opcl(f^∀[A])$.

The question: Is this co-continuity condition present somewhere in the literature? Does it have some nice properties? Or is there a reason why it is not interesting?

For now, I have made the following observations:

1. Co-continuous maps are stable under compositions, and similarly to continuity, it is easier to achieve if the domain topology is finer and if the co-domain topology is coarser.


2. The map $f: X to Y$ is co-continuous if for every open $U ⊆ f[X]$ and every $f$-section $s: U to X$ (i.e. $f ∘ s = opid_U$) the set $s[U]$ is also open.


3. In particular, if $f$ is co-continous, then $f^-1(y)$ is open discrete for every isolated point $y ∈ Y$.


4. If $f$ is co-continuous then $f^-1[U]$ is open for every open $U ⊆ f[X]$, so for surjective maps, co-continuity implies continuity. On the other hand, for injective maps, continuity implies co-continuity.


5. $f$ is co-continuous with respect to the topology $τ$ on $Y$ if and only if it is co-continuous with respect to $τ' := U ∈ τ: U ⊆ f[X] ∪ Y$. And $τ'$-co-continuity implies $τ'$-continuity. (Similarly, $τ$-continuity is equivalent to $τ''$-continuity where $τ''$ makes $f[X]$ open and every $y ∈ Y setminus f[X]$ isolated. Also, $τ''$-co-continuity implies $τ''$-continuity.)


6. In general, if $f[X]$ is open, then co-continuity implies continuity. On the other hand, if $f[X]$ has empty interior, then co-continuity is trivial.


7. If $f$ is co-continuous, then $f[C]$ is co-dense for every co-dense $C ⊆ X$, or equivalently $f^∀[D]$ is dense for every dense $D ⊆ X$.