1
$\begingroup$

Let (X, $\mathscr{T_x}$) and (Y, $\mathscr{T_y}$) be topological spaces. A map $f : X → Y$ is called continuous if the inverse image of each set open in $Y$ is open in X (that is $f^{−1}$ maps $\mathscr{T_y}$ into $\mathscr{T_x}$)

I want to understand what type of function is this if all this definition above

if the inverse image of each set open in Y is open in X

is also open in Y

In standard definition is continuous only if is open in X but it is not said that Y 'is opened' by continuity definition itself, i.e

{if the inverse image of each set open in Y is open in X} is opened in Y

In standard definition I see a closure, "the price to pay" is a 'closed Y' for a open X, not a opening Y from an opened X by Y

Y = the inverse image of each set open in $Y$ is open in X

In an equivalent way it is like saying that X is open by Y as inverse image of each set open in Y but this is a self-referential definition because is works like a closed cicle loop, only from Y to X, there is no return for both Y and X, if not exclusively for oneself: Y but not X

To have continuity the first subject is inverse image of each set open in Y (but Y can be 'closed') while X must be 'is open'.

This definition has only one verse, there is not a 'feedback' continuity in the "opposite" direction (which is not an opposite verse, but just an Y as open set respect to 'in Y is open in X')

I need a hetero-referential continuity (as a kind of a full duplex signal) not an self-referential continuity (as a half-duplex)

EDIT: open in Y is for me this

$\endgroup$

2 Answers 2

1
$\begingroup$

I don't really understand what the issue is. We have topological spaces $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$, and a function $f:X \to Y$.

$f$ is continuous iff

$$\forall O \in \mathscr{T}_Y: f^{-1}[O] \in \mathscr{T}_X$$

where as usual $f^{-1}[O] = \{x \in X: f(x) \in O\}$, so this has nothing to do with an inverse function, just a so-called inverse image.

So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $\mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $\mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.

$\endgroup$
1
0
$\begingroup$

I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".

You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .