Sequence covering map

In mathematics, specifically topology, a sequence covering map is any of a class of maps between topological spaces whose definitions all somehow relate sequences in the codomain with sequences in the domain. Examples include maps,, , and. These classes of maps are closely related to sequential spaces. If the domain and/or codomain have certain additional topological properties (often, the spaces being Hausdorff and first-countable is more than enough) then these definitions become equivalent to other well-known classes of maps, such as open maps or quotient maps, for example. In these situations, characterizations of such properties in terms of convergent sequences might provide benefits similar to those provided by, say for instance, the characterization of continuity in terms of sequential continuity or the characterization of compactness in terms of sequential compactness (whenever such characterizations hold).

Preliminaries
A subset $$S$$ of $$(X, \tau)$$ is said to be  if whenever a sequence in $$X$$ converges (in $$(X, \tau)$$) to some point that belongs to $$S,$$ then that sequence is necessarily in $$S$$ (i.e. at most finitely many points in the sequence do not belong to $$S$$). The set $$\operatorname{SeqOpen}(X, \tau)$$ of all sequentially open subsets of $$(X, \tau)$$ forms a topology on $$X$$ that is finer than $$X$$'s given topology $$\tau.$$ By definition, $$(X, \tau)$$ is called a  if $$\tau = \operatorname{SeqOpen}(X, \tau).$$ Given a sequence $$x_{\bull}$$ in $$X$$ and a point $$x \in X,$$ $$x_{\bull} \to x$$ in $$(X, \tau)$$ if and only if $$x_{\bull} \to x$$ in $$(X, \operatorname{SeqOpen}(X, \tau)).$$ Moreover, $$\operatorname{SeqOpen}(X, \tau)$$ is the topology on $$X$$ for which this characterization of sequence convergence in $$(X, \tau)$$ holds.

A map $$f : (X, \tau) \to (Y, \sigma)$$ is called  if $$f : (X, \operatorname{SeqOpen}(X, \tau)) \to (Y, \operatorname{SeqOpen}(Y, \sigma))$$ is continuous, which happens if and only if for every sequence $$x_{\bull} = \left(x_i\right)_{i=1}^{\infty}$$ in $$X$$ and every $$x \in X,$$ if $$x_{\bull} \to x$$ in $$(X, \tau)$$ then necessarily $$f\left(x_{\bull}\right) \to f(x)$$ in $$(Y, \sigma).$$ Every continuous map is sequentially continuous although in general, the converse may fail to hold. In fact, a space $$(X, \tau)$$ is a sequential space if and only if it has the following :
 * for every topological space $$(Y, \sigma)$$ and every map $$f : X \to Y,$$ the map $$f : (X, \tau) \to (Y, \sigma)$$ is continuous if and only if it is sequentially continuous.

The  in $$(X, \tau)$$ of a subset $$S \subseteq X$$ is the set $$\operatorname{scl}_{(X, \tau)} S$$ consisting of all $$x \in X$$ for which there exists a sequence in $$S$$ that converges to $$x$$ in $$(X, \tau).$$ A subset $$S \subseteq X$$ is called  in $$(X, \tau)$$ if $$S = \operatorname{scl}_{(X, \tau)} S,$$ which happens if and only if whenever a sequence in $$S$$ converges in $$(X, \tau)$$ to some point $$x \in X$$ then necessarily $$x \in S.$$ The space $$(X, \tau)$$ is called a  if $$\operatorname{scl}_X S ~=~ \operatorname{cl}_X S$$ for every subset $$S \subseteq X,$$ which happens if and only if every subspace of $$(X, \tau)$$ is a sequential space. Every first-countable space is a Fréchet–Urysohn space and thus also a sequential space. All pseudometrizable spaces, metrizable spaces, and second-countable spaces are first-countable.

Sequence coverings
A sequence $$x_{\bull} = \left(x_i\right)_{i=1}^\infty$$ in a set $$X$$ is by definition a function $$x_{\bull} : \N \to X$$ whose value at $$i \in \N$$ is denoted by $$x_i$$ (although the usual notation used with functions, such as parentheses $$x_{\bull}(i)$$ or composition $$f \circ x_{\bull},$$ might be used in certain situations to improve readability). Statements such as "the sequence $$x_{\bull}$$ is injective" or "the image (i.e. range) $$\operatorname{Im} x_{\bull}$$ of a sequence $$x_{\bull}$$ is infinite" as well as other terminology and notation that is defined for functions can thus be applied to sequences. A sequence $$s_{\bull}$$ is said to be a  of another sequence $$x_{\bull}$$ if there exists a strictly increasing map $$l_{\bull} : \N \to \N$$ (possibly denoted by $$l_{\bull} = \left(l_k\right)_{k=1}^{\infty}$$ instead) such that $$s_k = x_{l_k}$$ for every $$k \in \N,$$ where this condition can be expressed in terms of function composition $$\circ$$ as: $$s_{\bull} = x_{\bull} \circ l_{\bull}.$$ As usual, if $$x_{l_{\bull}} = \left(x_{l_k}\right)_{k=1}^{\infty}$$ is declared to be (such as by definition) a subsequence of $$x_{\bull}$$ then it should immediately be assumed that $$l_{\bull} : \N \to \N$$ is strictly increasing. The notation $$x_{\bull} \subseteq S$$ and $$\operatorname{Im} x_{\bull} \subseteq S$$ mean that the sequence $$x_{\bull}$$ is valued in the set $$S.$$

The function $$f : X \to Y$$ is called a  if for every convergent sequence $$y_{\bull}$$ in $$Y,$$ there exists a sequence $$x_{\bull} \subseteq X$$ such that $$y_{\bull} = f \circ x_{\bull}.$$ It is called a  if for every $$y \in Y$$ there exists some $$x \in f^{-1}(y)$$ such that every sequence $$y_{\bull} \subseteq Y$$ that converges to $$y$$ in $$(Y, \sigma),$$ there exists a sequence $$x_{\bull} \subseteq X$$ such that $$y_{\bull} = f \circ x_{\bull}$$ and $$x_{\bull}$$ converges to $$x$$ in $$(X, \tau).$$ It is a  if $$f : X \to Y$$ is surjective and also for every $$y \in Y$$ and every $$x \in f^{-1}(y),$$ every sequence $$y_{\bull} \subseteq Y$$ and converges to $$y$$ in $$(Y, \sigma),$$ there exists a sequence $$x_{\bull} \subseteq X$$ such that $$y_{\bull} = f \circ x_{\bull}$$ and $$x_{\bull}$$ converges to $$x$$ in $$(X, \tau).$$ A map $$f : X \to Y$$ is a  if for every compact $$K \subseteq Y$$ there exists some compact subset $$C \subseteq X$$ such that $$f(C) = K.$$

Sequentially quotient mappings
In analogy with the definition of sequential continuity, a map $$f : (X, \tau) \to (Y, \sigma)$$ is called a  if
 * $$f : (X, \operatorname{SeqOpen}(X, \tau)) \to (Y, \operatorname{SeqOpen}(Y, \sigma))$$

is a quotient map, which happens if and only if for any subset $$S \subseteq Y,$$ $$S$$ is sequentially open $$(Y, \sigma)$$ if and only if this is true of $$f^{-1}(S)$$ in $$(X, \tau).$$ Sequentially quotient maps were introduced in who defined them as above.

Every sequentially quotient map is necessarily surjective and sequentially continuous although they may fail to be continuous. If $$f : (X, \tau) \to (Y, \sigma)$$ is a sequentially continuous surjection whose domain $$(X, \tau)$$ is a sequential space, then $$f : (X, \tau) \to (Y, \sigma)$$ is a quotient map if and only if $$(Y, \sigma)$$ is a sequential space and $$f : (X, \tau) \to (Y, \sigma)$$ is a sequentially quotient map.

Call a space $$(Y, \sigma)$$  if $$(Y, \operatorname{SeqOpen}(Y, \sigma))$$ is a Hausdorff space. In an analogous manner, a "sequential version" of every other separation axiom can be defined in terms of whether or not the space $$(Y, \operatorname{SeqOpen}(Y, \sigma))$$ possess it. Every Hausdorff space is necessarily sequentially Hausdorff. A sequential space is Hausdorff if and only if it is sequentially Hausdorff.

If $$f : (X, \tau) \to (Y, \sigma)$$ is a sequentially continuous surjection then assuming that $$(Y, \sigma)$$ is sequentially Hausdorff, the following are equivalent:

 $$f : (X, \tau) \to (Y, \sigma)$$ is sequentially quotient. Whenever $$y_{\bull} \to y$$ is a convergent sequence in $$Y$$ then there exists a convergent sequence $$x_{\bull} \to x$$ in $$X$$ such that $$f(x) = y$$ and $$f \circ x_{\bull}$$ is a subsequence of $$y_{\bull}.$$  Whenever $$y_{\bull}$$ is a convergent sequence in $$Y$$ then there exists a convergent sequence $$x_{\bull}$$ in $$X$$ such that $$f \circ x_{\bull}$$ is a subsequence of $$y_{\bull}.$$  
 * This statement differs from (2) above only in that there are no requirements placed on the limits of the sequences (which becomes an important difference only when $$Y$$ is not sequentially Hausdorff).
 * If $$f : X \to Y$$ is a continuous surjection onto a sequentially compact space $$Y$$ then this condition holds even if $$Y$$ is not sequentially Hausdorff.

If the assumption that $$Y$$ is sequentially Hausdorff were to be removed, then statement (2) would still imply the other two statement but the above characterization would no longer be guaranteed to hold (however, if points in the codomain were required to be sequentially closed then any sequentially quotient map would necessarily satisfy condition (3)). This remains true even if the sequential continuity requirement on $$f : X \to Y$$ was strengthened to require (ordinary) continuity. Instead of using the original definition, some authors define "sequentially quotient map" to mean a surjection that satisfies condition (2) or alternatively, condition (3). If the codomain is sequentially Hausdorff then these definitions differs from the original in the added requirement of continuity (rather than merely requiring sequential continuity).

The map $$f : (X, \tau) \to (Y, \sigma)$$ is called  if for every convergent sequence $$y_{\bull} \to y$$ in $$(Y, \sigma)$$ such that $$y_{\bull}$$ is not eventually equal to $$y,$$ the set $$\bigcup_{\stackrel{i \in \N,}{y_i \neq y}} f^{-1}\left(y_i\right)$$ is sequentially closed in $$(X, \tau),$$ where this set may also be described as:
 * $$\bigcup_{\stackrel{i \in \N,}{y_i \neq y}} f^{-1}\left(y_i\right)

~=~ f^{-1} \left(\left(\operatorname{Im} y_{\bull}\right) \setminus \{ y \}\right) ~=~ f^{-1} \left(\operatorname{Im} y_{\bull}\right) \setminus f^{-1}(y) $$ Equivalently, $$f : (X, \tau) \to (Y, \sigma)$$ is presequential if and only if for every convergent sequence $$y_{\bull} \to y$$ in $$(Y, \sigma)$$ such that $$y_{\bull} \subseteq Y \setminus \{ y \},$$ the set $$f^{-1} \left(\operatorname{Im} y_{\bull}\right)$$ is sequentially closed in $$(X, \tau).$$

A surjective map $$f : (X, \tau) \to (Y, \sigma)$$ between Hausdorff spaces is sequentially quotient if and only if it is sequentially continuous and a presequential map.

Characterizations
If $$f : (X, \tau) \to (Y, \sigma)$$ is a continuous surjection between two first-countable Hausdorff spaces then the following statements are true:  $$f$$ is almost open if and only if it is a 1-sequence covering. $$f$$ is an open map if and only if it is a 2-sequence covering. If $$f$$ is a compact covering map then $$f$$ is a quotient map. The following are equivalent:  $$f$$ is a quotient map.</li> $$f$$ is a sequentially quotient map.</li> $$f$$ is a sequence covering.</li> $$f$$ is a pseudo-open map. </ol> and if in addition both $$X$$ and $$Y$$ are separable metric spaces then to this list may be appended:  $$f$$ is a hereditarily quotient map.</li> </ol> </li> </ul>
 * An ' is surjective map $$f : X \to Y$$ with the property that for every $$y \in Y,$$ there exists some $$x \in f^{-1}(y)$$ such that $$x$$ is a ' for $$f,$$ which by definition means that for every open neighborhood $$U$$ of $$x,$$ $$f(U)$$ is a neighborhood of $$f(x)$$ in $$Y.$$</li>
 * A map $$f : X \to Y$$ is called  if for every $$y \in Y$$ and every open neighborhood $$U$$ of $$f^{-1}(y)$$ (meaning an open subset $$U$$ such that $$f^{-1}(y) \subseteq U$$), $$y$$ necessarily belongs to the interior (taken in $$Y$$) of $$f(U).$$</li>

Properties
The following is a sufficient condition for a continuous surjection to be sequentially open, which with additional assumptions, results in a characterization of open maps. Assume that $$f : X \to Y$$ is a continuous surjection from a regular space $$X$$ onto a Hausdorff space $$Y.$$ If the restriction $$f\big\vert_U : U \to f(U)$$ is sequentially quotient for every open subset $$U$$ of $$X$$ then $$f : X \to Y$$ maps open subsets of $$X$$ to sequentially open subsets of $$Y.$$ Consequently, if $$X$$ and $$Y$$ are also sequential spaces, then $$f : X \to Y$$ is an open map if and only if $$f\big\vert_U : U \to f(U)$$ is sequentially quotient (or equivalently, quotient) for every open subset $$U$$ of $$X.$$

Given an element $$y \in Y$$ in the codomain of a (not necessarily surjective) continuous function $$f : X \to Y,$$ the following gives a sufficient condition for $$y$$ to belong to $$f$$'s image: $$y \in \operatorname{Im} f := f(X).$$ A family $$\mathcal{B}$$ of subsets of a topological space $$(X, \tau)$$ is said to be  at a point $$x \in X$$ if there exists some open neighborhood $$U$$ of $$x$$ such that the set $$\left\{ B \in \mathcal{B} ~:~ U \cap B \neq \varnothing \right\}$$ is finite. Assume that $$f : X \to Y$$ is a continuous map between two Hausdorff first-countable spaces and let $$y \in Y.$$ If there exists a sequence $$y_{\bull} = \left(y_i\right)_{i=1}^{\infty}$$ in $$Y$$ such that (1) $$y_{\bull} \to y$$ and (2) there exists some $$x \in X$$ such that $$\left\{ f^{-1}\left(y_i\right) ~:~ i \in \N \right\}$$ is locally finite at $$x,$$ then $$y \in \operatorname{Im} f = f(X).$$ The converse is true if there is no point at which $$f$$ is locally constant; that is, if there does not exist any non-empty open subset of $$X$$ on which $$f$$ restricts to a constant map.

Sufficient conditions
Suppose $$f : X \to Y$$ is a continuous open surjection from a first-countable space $$X$$ onto a Hausdorff space $$Y,$$ let $$D \subseteq Y$$ be any non-empty subset, and let $$y \in \operatorname{cl}_Y D$$ where $$\operatorname{cl}_Y D$$ denotes the closure of $$D$$ in $$Y.$$ Then given any $$x, z \in f^{-1}(y)$$ and any sequence $$x_{\bull}$$ in $$f^{-1}(D)$$ that converges to $$x,$$ there exists a sequence $$z_{\bull}$$ in $$f^{-1}(D)$$ that converges to $$z$$ as well as a subsequence $$\left(x_{l_k}\right)_{k=1}^\infty$$ of $$x_{\bull}$$ such that $$f(z_k) = f\left(x_{l_k}\right)$$ for all $$k \in \N.$$ In short, this states that given a convergent sequence $$x_{\bull} \subseteq f^{-1}(D)$$ such that $$x_{\bull} \to x$$ then for any other $$z \in f^{-1}(f(x))$$ belonging to the same fiber as $$x,$$ it is always possible to find a subsequence $$x_{l_{\bull}} = \left(x_{l_k}\right)_{k=1}^\infty$$ such that $$f \circ x_{l_{\bull}} = \left(f\left(x_{l_k}\right)\right)_{k=1}^\infty$$ can be "lifted" by $$f$$ to a sequence that converges to $$z.$$

The following shows that under certain conditions, a map's fiber being a countable set is enough to guarantee the existence of a point of openness. If $$f : X \to Y$$ is a sequence covering from a Hausdorff sequential space $$X$$ onto a Hausdorff first-countable space $$Y$$ and if $$y \in Y$$ is such that the fiber $$f^{-1}(y)$$ is a countable set, then there exists some $$x \in f^{-1}(y)$$ such that $$x$$ is a point of openness for $$f : X \to Y.$$ Consequently, if $$f : X \to Y$$ is quotient map between two Hausdorff first-countable spaces and if every fiber of $$f$$ is countable, then $$f : X \to Y$$ is an almost open map and consequently, also a 1-sequence covering.