# Continuous map

(Redirected from Homomorphism (topology))
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
Note: there are a few different conditions for continuity, there's also continuity at a point. This diagram is supposed to show how they relate to each other.
 Overview Key $\begin{xy}\xymatrix{ \text{Continuous} \ar@2{<->}[d]_-{\text{claim }1} \ar@2{<.>}[drr] & & \\ {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(neighbourhood)}\end{array} } \ar@2{<->}[rr]_{\text{claim }2} & & {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(sequential)}\end{array} } }\end{xy}$ Note that: All arrow denote logical implies, or "if and only if" Dotted arrows show immediate results of the claims on this page

## Definition

Given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] we say that a map, [ilmath]f:X\rightarrow Y[/ilmath] is continuous if:

• [ilmath]\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}][/ilmath]

That is to say:

• The pre-image of every set open in [ilmath]Y[/ilmath] under [ilmath]f[/ilmath] is open in [ilmath]X[/ilmath]

## Continuous at a point

Again, given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath], and a point [ilmath]x_0\in X[/ilmath], we say the map [ilmath]f:X\rightarrow Y[/ilmath] is continuous at [ilmath]x_0[/ilmath] if:

### Claim 1

Claim: The mapping [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] it is continuous at every point

• A quick proof I did on some scrap - click for full version

This requires one or more proofs to be written up neatly and is on a to-do list for having them written up. This does not mean the results cannot be trusted, it means the proof has been completed, just not written up here yet. It may be in a notebook, some notes about reproducing it may be left in its place, perhaps a picture of it, so forth. The message provided is:
See image

## Sequentially continuous at a point

Given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath], and a point [ilmath]x_0\in X[/ilmath], a function [ilmath]f:X\rightarrow Y[/ilmath] is said to be continuous at [ilmath]x_0[/ilmath] if:

• $\forall (x_n)_{n=1}^\infty\left[\lim_{n\rightarrow\infty}(x_n)=x\implies\lim_{n\rightarrow\infty}(f(x_n))=f(x)\right]$ (Recall that [ilmath](x_n)_{n=1}^\infty[/ilmath] denotes a sequence, see Limit (sequence) for information on limits)

### Claim 2

Claim: [ilmath]f[/ilmath] is continuous at [ilmath]x_0[/ilmath] using the neighbourhood definition [ilmath]\iff[/ilmath] it is continuous at [ilmath]x_0[/ilmath] using the sequential definition

Proof: neighbourhood [ilmath]\implies[/ilmath] sequential:

Let [ilmath](x_n)_{n=1}^\infty[/ilmath] be given, and that [ilmath](x_n)\rightarrow x[/ilmath] - we wish to show that [ilmath]\lim_{n\rightarrow\infty}(f(x_n))=f(x)[/ilmath]
Let [ilmath]\epsilon > 0[/ilmath] be given.
By hypothesis, as [ilmath]B_\epsilon(f(x))[/ilmath] is a neighbourhood of [ilmath]f(x)[/ilmath] then [ilmath]f^{-1}(B_\epsilon(f(x)))[/ilmath] is a neighbourhood to [ilmath]x[/ilmath]
So [ilmath]\exists\delta>0[B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))[/ilmath]
• By the implies-subset relation if [ilmath]B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))[/ilmath] then [ilmath]a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))[/ilmath]
Choose [ilmath]N\in\mathbb{N} [/ilmath] such that [ilmath]n>N\implies d_1(x_n,x)<\delta[/ilmath] (which we can do because [ilmath](x_n)\rightarrow x[/ilmath])
• Note that [ilmath]d_1(x_n,x)<\delta\implies x_n\in B_\delta(x)[/ilmath]
So [ilmath]x_n\in B_\delta(x)\implies x_n\in f^{-1}(B_\epsilon(f(x)))[/ilmath]
• But if [ilmath]a\in f^{-1}(B)[/ilmath] then [ilmath]f(a)\in B[/ilmath] as [ilmath]f^{-1}(B)[/ilmath] contains exactly the things which map to an element of [ilmath]B[/ilmath] under [ilmath]f[/ilmath]
So [ilmath]f(x_n)\in B_\epsilon(f(x))[/ilmath]
But [ilmath]f(x_n)\in B_\epsilon(f(x))\iff d_2(f(x_n),f(x))<\epsilon[/ilmath]
This completes the proof
• we have shown that given an [ilmath]\epsilon > 0[/ilmath] that there exists an [ilmath]N[/ilmath] such that for [ilmath]n>N[/ilmath] we have [ilmath]d_2(f(x_n),f(x))<\epsilon[/ilmath] which is exactly the definition of [ilmath](f(x_n))\rightarrow f(x)[/ilmath]

Proof: sequential [ilmath]\implies[/ilmath] neighbourhood:

TODO: This - See Analysis I - Maurin page 44 if stuck