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So I understand that for a "continuous" map between topological spaces, the pre-image of an open subset is an open subset. (There is no 'inverse map' unless the function is bijective, so careful there)

But in general...when one just says "map" between $X$ and $Y$, (though sometimes it's implicitly intended that it is continuous, in cases otherwise), is there a rule?

Like say, a map $\phi:X \to Y$ always maps an open element to an open element? I ask this because for a "continuous" map, the "inverse" is forced to map an open element from $Y$ to an open element of $X$. But nothing is said about $\phi$ itself. Does $\phi$ have to map an open element from $X$ to an open element of $Y$?

Or is this automatic if the inverse maps an open element to an open element? It seems so to me, but not very sure...though, being an "inverse" it logically makes sense I think that the statement is implied.

But I wish to have some clarification(I don't want to assume things and go on and realise I had a logic hole somewhere later)

...Actually, when we "map" topological space to another, are we mapping "sets"? Regardless of their size(cardinality)? So say, if $X=\{0,1\}$ and the topology $\tau=\{\phi,X,\{0\}\}$ is together a topological space $X$, and $Y=\{0\}$, then can I map $\{0,1\} \to \{0\}$? The former has cardinality $2$ and the latter $1$ but does this not matter? I'm confused since the "map" is slightly different from conventional "functions" that map elements to elements...I guess $\{0\}$ is not an "element" but a singleton "set" in a topological context?

Melba1993
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    Wait, what is an "open element" ? If you mean an open subset, then maps that do take an open subset to an open subset are called "open maps". See https://en.wikipedia.org/wiki/Open_and_closed_maps – Captain Lama Mar 29 '16 at 19:40
  • Actually, let me update my question since that confuses me haha – Melba1993 Mar 29 '16 at 19:45
  • Map is usually a synonym for function. Anything can be a member of a set, including another set. Member is a synonym for element. A map $f:X \to Y $("from$ X$ to $Y$") is any $ f\subset {(p,q):p\in X\land q\in Y}$ such that for each $p\in X$ there is exactly one $q\in Y$ such that $(p,q)\in f.$ When $(p,q)\in f$ we write $q=f(p).$ Note that this defines a function as being equal to "its graph." (Well, how else would you define it?) – DanielWainfleet Mar 30 '16 at 01:43
  • if you find DanielWainfleet's comment interesting: https://math.stackexchange.com/questions/95741/is-there-any-difference-between-mapping-and-function – Pineapple Fish Dec 08 '23 at 05:56

2 Answers2

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1) For starters, a map between metric spaces, $f:X\to Y$, is continuous if and only if the pre-image $f^{-1}(V)$ of any open subset $V\subset Y$ is itself an open subset of $X$.

Since it suffices to check this for balls, taking $V=B(f(x_0),\epsilon)$ we obtain that the ($\epsilon-\delta$) criterion for $f$ around $x_0$ is equivalent to $f^{-1}(V)$ containing a ball centred at $x_0.$

2) The way textbooks are organised generally follows the tidy structure first laid out by the Bourbaki group. That is to say, when we are in the category (=context, for now) of differentiable manifolds, we assume that a map between such manifolds is differentiable; when the background is that of topological spaces, maps are continuous -- or else you are ignoring the topology, and simply stripping the spaces of it and merely viewing as sets!

3) You are making reference to $f$(open subset) possibly being open, but this is a different condition. It suffices to consider, say, the following function $$f:X\to X\times \mathbb{R},$$ defined by $f(x)=(x,0)$. All open sets on $X\times \mathbb{R}$ are unions of products $U\times V$ where $U$ is open in $X$ and $V$ is open in $\mathbb{R}$, so obviously there is no open subset of $X$ whose image is open in the target space.

Theon Alexander
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  • Hi, I'm baffled with small questions like, in topological spaces, does some map $\phi$(empty set) to another empty set? Or some subset to an empty set? – Melba1993 Mar 29 '16 at 20:16
  • A topological space is a set plus extra structure, therefore maps of topological spaces, i.e. continuous maps, are maps of sets, and the image of the empty set is empty (just imagine an empty diagram). – Theon Alexander Mar 29 '16 at 20:41
  • You may enjoy this e-book http://www.topologywithouttears.net/topbook.pdf – Theon Alexander Mar 29 '16 at 20:46
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A map does not have to map open sets to open sets.

One example is the continuous constant maps $f : X \to \mathbb{R}$.

It maps open sets to closed sets.

JKnecht
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