Operator characterizations of continuity and “co-continuity”

Question

$\let \op \operatorname$There are several characterizations of continuity of a map $f: X \to Y$ between topological spaces of the form $f^•[\op{op} A] \diamond \op{op} f^•[A]$ for every $A ⊆ X$ or $Y$ where $f^• ∈ \{f, f^{-1}, f^∀\}$ denotes image, preimage, and dual image ($f^∀[A] := Y \setminus f[X \setminus A]$), $\op{op} ∈ \{\op{cl}, \op{int}\}$ 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 $F ⊆ Y$:

• $f[\op{cl}(A)] ⊆ \op{cl}(f[A])$,
• $f^∀[\op{int}(A)] ⊇ \op{int}(f^∀[A])$,
• $f^{-1}[\op{cl}(B)] ⊇ \op{cl}(f^{-1}[B])$,
• $f^{-1}[\op{int}(B)] ⊆ \op{int}(f^{-1}[B])$.

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 $U ⊆ X$:

• $f[\op{cl}(A)] ⊇ \op{cl}(f[A])$,
• $f^∀[\op{int}(A)] ⊆ \op{int}(f^∀[A])$.

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 $F ⊆ X$:

• $f[\op{int}(A)] ⊆ \op{int}(f[A])$,
• $f^∀[\op{cl}(A)] ⊇ \op{cl}(f^∀[A])$,
• $f^{-1}[\op{int}(B)] ⊇ \op{int}(f^{-1}[B])$,
• $f^{-1}[\op{cl}(B)] ⊆ \op{cl}(f^{-1}[B])$.

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

• $f[\op{int}(A)] ⊇ \op{int}(f[A])$,
• $f^∀[\op{cl}(A)] ⊆ \op{cl}(f^∀[A])$.

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 (in $Y$) $U ⊆ f[X]$ and every $f$-section $s: U \to X$ (i.e. $f ∘ s = \op{id}_U$) 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 (in $Y$) $U ⊆ f[X]$, 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 $D ⊆ X$.