Open and closed maps
Due to the parallel definitions and other similarities there is little point to having separate pages.
Contents
Definition
Open map
A [math]f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})[/math] (which need not be continuous) is said to be an open map if:
- The image of an open set is open (that is [math]\forall U\in\mathcal{J}[f(U)\in\mathcal{K}][/math])
Closed map
A [math]f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})[/math] (which need not be continuous) is said to be a closed map if:
- The image of a closed set is closed
TODO: References - it'd look better
Importance with respect to the quotient topology
The primary use of recognising an open/closed map comes from recognising a Quotient topology, as the following theorem shows
Theorem: Given a map [math]f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/math] that is continuous, surjective and an open or closed map, then the topology [ilmath]\mathcal{K} [/ilmath] on [ilmath]Y[/ilmath] is the same as the Quotient topology induced on [ilmath]Y[/ilmath] by [ilmath]f[/ilmath]
Note: the intuition behind this theorem is fairly obvious, if [ilmath]f[/ilmath] is an open map then it takes open sets to open sets, but the quotient topology is precisely those sets in [ilmath]Y[/ilmath] whose preimage is open in [ilmath]X[/ilmath]
Using the first theorem, automatically we see that [ilmath]K[/ilmath] is weaker than the quotient topology (that is [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath])
To show equality we need only to show [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]
Open case:
- Suppose that [ilmath]f[/ilmath] is a continuous open map
- Let [ilmath]A\in\mathcal{Q} [/ilmath] be given
- we need to show this [ilmath]\implies A\in\mathcal{K} [/ilmath] then we use the Implies and subset relation to get what we need.
- This means [ilmath]f^{-1}(A)\in\mathcal{J} [/ilmath] but as [ilmath]f[/ilmath] is an open mapping [ilmath]\forall U\in\mathcal{J}[f(U)\in\mathcal{K}][/ilmath] so:
- [ilmath]f(f^{-1}(A))\in\mathcal{K} [/ilmath] (it is open in [ilmath]Y[/ilmath])
- Warning: we have not shown that [ilmath]A\in\mathcal{K} [/ilmath] yet!
- Because [ilmath]f[/ilmath] is surjective we know [ilmath]f(f^{-1}(A))=A[/ilmath] (as for all things in [ilmath]A[/ilmath] there must be something that maps to them, by surjectivity)
- So we conclude [ilmath]A\in \mathcal{K} [/ilmath]
- So [ilmath]A\in\mathcal{Q}\implies A\in\mathcal{K} [/ilmath] or [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]
- Combining this with [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath] we see:
- [ilmath]\mathcal{K}=\mathcal{Q}[/ilmath]
Closed case:
- Suppose [ilmath]f[/ilmath] is a continuous closed map
- Let [ilmath]A\in\mathcal{Q} [/ilmath] be given
- we need to show this [ilmath]\implies A\in\mathcal{K} [/ilmath] then we use the Implies and subset relation to get what we need.
- Obviously, because [ilmath]f[/ilmath] is continuous [ilmath]f^{-1}(A)\in\mathcal{J} [/ilmath] - that is to say [ilmath]f^{-1}(A)[/ilmath] is open in [ilmath]X[/ilmath]
- That means that [ilmath]X-f^{-1}(A)[/ilmath] is closed in [ilmath]X[/ilmath] - just by definition of an open set
- Since [ilmath]f[/ilmath] is a closed map, [ilmath]f(X-f^{-1}(A))[/ilmath] is also closed
- BUT! [ilmath]f(X-f^{-1}(A))=Y-A[/ilmath] by surjectivity (as the set [ilmath]X[/ilmath] take away ALL the things that map to points in [ilmath]A[/ilmath] will give you all of [ilmath]Y[/ilmath] less the points in [ilmath]A[/ilmath])
- As [ilmath]f(X-f^{-1}(A))[/ilmath] is closed and equal to [ilmath]Y-A[/ilmath], [ilmath]Y-A[/ilmath] is closed
- this means [ilmath]Y-(Y-A)[/ilmath] is open, but [ilmath]Y-(Y-A)=A[/ilmath] so [ilmath]A[/ilmath] is open.
- That means [ilmath]A\in\mathcal{K} [/ilmath]
- We have shown [ilmath]A\in\mathcal{Q}\implies A\in\mathcal{K} [/ilmath] and again by the Implies and subset relation we see [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]
- Combining this with [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath] we see:
- [ilmath]\mathcal{K}=\mathcal{Q}[/ilmath]
This completes the proof.
