Image of a Topologically Regular Set? This question has come up as a small question in my research and I think I'm a little too thick in the weeds with extraneous details to see it cleanly. If necessary, I can add some more conditions on the spaces and maps that follow---just ask!
Suppose $(X, d_x, \mu_x)$ and $(Y, d_y, \mu_y)$ are metric measure spaces, with metrics $d_x$ and $d_y$ and Borel measures $\mu_x$ and $\mu_y$ respectively. Let $f:X \to Y$ be continuous (and hence measureable). Generally assume that $X$ and $Y$ are "nice" (e.g., manifolds) but $f$ is "messy."

Definition: A set $U$ is called topologically regular (or a regular closed set) if it is the closure of its interior, i.e., $\overline{\mathrm{int } U} = U$.
Theorem: If $A \subseteq X$ is compact and $f$ is continous, then $f(A) \subseteq Y$ is compact.

Suppose for everything that follows that $A \subseteq X$ is non-empty, compact, and topologically regular. My aim is to establish the weakest possible condition on $f$ such that $f(A) \subseteq Y$ is also topologically regular.
Baby Question: If $f:X \to Y$ is continuous, is $f(A) \subseteq Y$ topologically regular?
Counter Example: Consider $f:\mathbb{R}^2 \to \mathbb{R}^2$ with the Euclidean metrics and Lebesgue measures. Let $f(x,y):=(x, 0)$; observe that $f$ is clearly continuous. However, $[0,1]^2$ is topologically regular but $f([0,1]^2) = [0,1] \times \{0\}$ which has empty interior and is hence not topologically regular. $\blacksquare$
Clearly we need a stronger condition on $f$ in order to guarantee that the image is also topologically regular. The following would be sufficient, but is a stronger condition than I could ever possibly dream of in my context.

Theorem: If $f:X \to Y$ is a homeomorphism, then $f(A)$ is topologically regular.

Proof: Clearly $f( \mathrm{int}~A) = \mathrm{int}~ f(A)$ as $f$ is a homeomorphism; since $A$ is topologically regular and non-empty, we have that $\mathrm{int}~A \neq \emptyset$ and hence $\mathrm{int}~f(A) \neq \emptyset$.
Let $y \in \partial f(A)$ and $\epsilon > 0$ be given. We seek to show that the ball $B(y, \epsilon)$ has non-trivial intersection with $\mathrm{int}~f(A)$.  Since $f(A)$ is compact (and hence closed), we have that $y \in f(A)$ and hence $x:= f^{-1}(y) \in A$. Since $A$ is topologically regular, we have that $f^{-1}(B(y, \epsilon)) \cap \mathrm{int}~A \neq \emptyset$ and therefore
$$B(y, \epsilon) \cap f(\mathrm{int}~A) = B(y, \epsilon) \cap \mathrm{int}~f(A) \neq \emptyset.$$
Hence $y \in \overline{\mathrm{int}~f(A)}$ and thus $f(A) = \overline{\mathrm{int}~f(A)}$ as desired. $\blacksquare$
The Question
In my specific context, $f$ will land somewhere on the spectrum between being continuous (trivially easy) and being a homeomorphism (provably impossible). What are the weakest possible conditions on $f$ to guarantee that $f(A)$ is topologically regular? Do any (or all?) of the following suffice? If they fail, what additional (if any) hypotheses on $f$ or the spaces $X$ and $Y$ are necessary?

*

*$f:X \to Y$ is continuous and bounded-to-one, i.e., there exists $M \in \mathbb{N}$ such that $\mathrm{Card } f^{-1}(y) \leq M$ for all $y \in Y$.


*$f:X \to Y$ is continuous and almost everywhere constant-to-one, i.e., there exists $M \in \mathbb{N}$ such that $\mathrm{Card}~f^{-1}(y) = M$ almost everywhere.


*$f:X \to Y$ is continuous and almost everywhere injective.


*$f: X \to Y$ is Lipschitz continuous. Edit: See above counter example


*$f: X \to Y$ is $\alpha$-Holder continuous. Edit: See above counter example
Edit
Per comments from @WilliamElliot and @HennoBrandsma, $f$ being continuous and an open (or closed) mapping suffices. The proof is similar to that of $f$ being a homeomorphism as above, but with some care taken as to inclusions instead of inequalities. In my specific context, this actually solves my problem as I can show that $f$ is a closed map. I still think there's some interesting analysis to be done regarding the five conditions I've listed above, so I'll leave the question open for now.
Edit 2: Open Mapping
There seems to be some confusion in the comments. I give a proof below that $f$ is continuous and an open mapping suffices. I also give proof that my Counter-Example is neither an open mapping nor a closed mapping.

Proposition E1: The map $f:\mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y):= (x,0)$ is not an open mapping.

Proof: Given open $U \subseteq \mathbb{R}^2$, we have that $f(U)= A \times \{0\}$ for some $A \subseteq \mathbb{R}$, which is not open in $\mathbb{R}^2$. $\blacksquare$

Proposition E2: The map $f:\mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y):= (x,0)$ is not a closed mapping.

Proof. Define $U:=\{(x,y) \in \mathbb{R}^2 \mid y \geq 1/x \text{ and } x > 0\}$. We have that $U$ is closed, but $f(U) = (0, \infty)\times \{0\}$ which is not closed. $\blacksquare$

Proposition E3: Suppose $f:X \to Y$ is continuous. If $f$ is an open mapping and $A \subseteq X$ is topologically regular, then $f(A)$ is topologically regular.

Proof.
Let topologically regular $A \subseteq X$ be given.
We begin by noting that if $A=\emptyset$, then the proposition holds vacuously. Therefore, assume that $A \neq \emptyset$ and thus $\mathrm{int} A \neq \emptyset$. Since $\mathrm{int} A \subseteq A$, we have that $f(\mathrm{int} A) \subseteq \mathrm{int} f(A)$ and in particular is non-empty.
Let $y \in f(A)$ and open neighborhood $U\ni y$ be given. Since $f$ is continuous, $f^{-1}(U)$ is open in $X$ and in particular $f^{-1}(U) \cap A$ is non-empty. As $\overline{\mathrm{int} A} = A$, we have that $f^{-1}(U) \cap \mathrm{int} A \neq \emptyset$. Therefore
$$U \cap \mathrm{int} f(A) \supseteq U \cap f(\mathrm{int} A) \supseteq f \left(f^{-1}(U) \cap \mathrm{int} A \right) \neq \emptyset$$
and thus $y \in \overline{\mathrm{int} f(A)}$ as desired.
$\blacksquare$
Edit 3: Closed Mapping
Having $f$ be continuous and a closed mapping is insufficient to guarantee that the image is topologically regular.
Counter-Example: Consider $f:[0,1]^2 \to [0,1]^2$ (with the standard topologies) defined by $f(x,y):=(x,0)$. We can clearly see that $f$ is continuous and we will show that $f$ is a closed mapping. Let a closed subset $A \subseteq [0,1]^2$. As $A$ is a closed and bounded subset of $\mathbb{R}^2$, we have that $A$ is compact. Since the image of a compact set is compact (and thus closed), we have that $f(A)$ is closed as well and thus $f$ is a closed mapping.
However, $f(A) \subseteq [0,1] \times \{0\}$ and thus $\mathrm{int}~f(A) = \emptyset$ in $[0,1]^2$. Therefore, given any non-empty $A$ we have that $\overline{\mathrm{int} f(A)} = \emptyset \neq f(A)$ and thus $f(A)$ cannot be topologically regular. $\blacksquare$
 A: We give a counter-example which shows that none of the following conditions are sufficient to guarantee that the image of a regular set is regular.  More conditions on $X,Y, f$ could presumably be added so as to give sufficiency, but this vague enough as to not be a well-posed question. 


*

*$f:X \to Y$ is continuous and bounded-to-one, i.e., there exists $M \in \mathbb{N}$ such that $\mathrm{Card } f^{-1}(y) \leq M$ for all $y \in Y$.

*$f:X \to Y$ is continuous and almost everywhere constant-to-one, i.e., there exists $M \in \mathbb{N}$ such that $\mathrm{Card}~f^{-1}(y) = M$ almost everywhere.

*$f:X \to Y$ is continuous and almost everywhere injective.

*$f: X \to Y$ is continuous and injective.
Counter-Example: Let $f: \{0\} \to \mathbb{R}$ be defined by $f(0):=0$. Equip $\mathbb{R}$ with the standard topology, Euclidean metric, and Lebesgue measure $\mu$. Equip $\{0\}$ with the discrete topology (which is the same as the trivial topology in this case), the discrete metric, and the discrete measure $\lambda$. Observe the following:


*

*The Lebesgue measure $\mu$ is a Borel measure on $\mathbb{R}$.

*$\lambda$ is a Borel measure on $\{0\}$: the only open sets are $\emptyset$ and $\{0\}$, which have measures $0$ and $1$ respectively. Hence every open set is measurable and thus $\lambda$ is a Borel measure. 

*$f$ is continuous: Let open $U \subseteq \mathbb{R}$ be given. Either $U$ contains $0$ or it doesn't. If $0 \in U $, then $f^{-1}(U) = \{0\}$ which is open. If $0 \not \in U $, then $f^{-1}(U) = \emptyset$ which is also open.  Therefore the preimage of any open set is open and thus $f$ is continuous. 

*$f$ is Lipschitz continuous and Holder continuous: trivial. 

*$f$ is injective: suppose $f(x) = f(x')$. Since $\{0\}$ contains a single point, we have that $x=0=x'$ and thus $f$ is injective.

*$f$ is almost everywhere injective: this follows immediately from $f$ being injective. Explicitly, let $N \subseteq \{0\}$ be the set on which $f$ is not injective. Then $N=\emptyset$ so $\lambda(N) = \lambda(\emptyset) = 0$ as desired.

*$f$ is bounded-to-one: given $y \in \mathbb{R}$, we have that $\mathrm{Card}~f^{-1}(y)$ is either $0$ or $1$. Hence $f$ is bounded-to-one.

*$f$ is almost everywhere constant-to-one: this follows immediately from $f$ being injective or from $f$ being almost everywhere injective. 

*$f$ is a closed mapping: we have that $f(\emptyset) = \emptyset$ and $f(\{0\}) = \{0\}$, which are both closed in $\mathbb{R}$. Hence the image of every closed subset of $\{0\}$ is closed in $\mathbb{R}$. 

*$f$ is NOT an open mapping: $\{0\}$ is an open subset of $\{0\}$, but $f(\{0\}) = \{0\}$ is not open in $\mathbb{R}$. 


Observe that $\{0\}$ is a regular closed set in $\{0\}$. Explicitly, $\{0\}$ is equipped with the discrete topology, so $\{0\}$ is open and thus $\mathrm{int}~\{0\} = \{0\}$. Furthermore, $\{0\}$ is closed and thus $\overline{\mathrm{int}~\{0\}} = \{0\}$. 
However, $f(\{0\}) = \{0\} \subseteq \mathbb{R}$ has empty interior and thus $\overline{\mathrm{int}~f(\{0\})} = \emptyset \neq \{0\} = f(\{0\})$. Therefore $f$ does not have the property of mapping regular closed sets to regular closed sets as desired. $\blacksquare$
