Difference between revisions of "Exercises:Mond - Topology - 1/Question 8"
(Created page with "<noinclude> ==Section B== ===Question 8=== </noinclude>Suppose that {{M|f:X\rightarrow Y}} is a continuous function that is also surjective, and let {{M|\sim}} denote...") |
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Recall the result of question 5: | Recall the result of question 5: | ||
* "''[[If a surjective continuous map is factored through the canonical projection of the equivalence relation induced by that map then the yielded map is a continuous bijection]]''" | * "''[[If a surjective continuous map is factored through the canonical projection of the equivalence relation induced by that map then the yielded map is a continuous bijection]]''" | ||
| − | Using | + | Using the linked theorem we instantly obtain: |
* {{M|\bar{f}:\frac{X}{\sim}\rightarrow Y}} that is a ''[[continuous]]'' [[bijection]] | * {{M|\bar{f}:\frac{X}{\sim}\rightarrow Y}} that is a ''[[continuous]]'' [[bijection]] | ||
** We must show this is a [[homeomorphism]]. | ** We must show this is a [[homeomorphism]]. | ||
| − | Suppose {{M|X}} is [[compact]], then {{M|\frac{X}{\sim} }} is compact also as [[the image of a compact set is compact]] | + | Suppose {{M|X}} is [[compact]], then {{M|\frac{X}{\sim} }} is compact also as [[the image of a compact set is compact]] and {{M|\pi:X\rightarrow\frac{X}{\sim} }} is continuous (see [[quotient topology]] for more details) |
| − | + | ||
We then apply the "''[[compact-to-Hausdorff theorem]]''" which shows us that {{M|\bar{f} }} is actually a [[homeomorphism]] | We then apply the "''[[compact-to-Hausdorff theorem]]''" which shows us that {{M|\bar{f} }} is actually a [[homeomorphism]] | ||
| − | This is similar to the [[first group isomorphism theorem]] in | + | This is similar to the ''[[first group isomorphism theorem]]'' (in the case the map is surjective) certainly in a [[Category Theory (subject)|categorical]] sense. The group theorem "factors through the kernel of the morphism" where as this "factors through the equivalence relation induced by the morphism" and they both yield an isomorphism (a homeomorphism is the term for a topological [[isomorphism (category theory)|isomorphism]]. Here are diagrams: |
| + | * For any ([[surjective]]) [[group homomorphism]] {{M|\varphi:G\rightarrow H}} (for arbitrary groups) we get a [[group isomorphism]], {{M|\bar{\varphi} }}. {{M|1=\xymatrix{ G \ar[r]^\varphi \ar[d]_\pi & H \\ {G/\text{Ker}(\varphi)} \ar@{.>}[ur]_{\bar{\varphi} } } }}, and, | ||
| + | * For any ([[surjective]]) [[continuous map]], {{M|f:X\rightarrow Y}} (for an arbitrary compact space {{M|X}}, and arbitrary [[Hausdorff]] space {{M|Y}}) we get a homeomorphism/topological isomorphism {{M|\bar{f} }}. {{M|1=\xymatrix{ X \ar[d]_\pi \ar[r]^f & Y \\ \frac{X}{x_1\sim x_2\iff f(x_1)=f(x_2)} \ar@{.>}[ur]_(.6){\bar{f} } } }} | ||
| + | In both instances the "denominator" of the quotient depends on the map, factoring (as in [[passing to the quotient (function)]] is involved, rather than [[passing to the quotient (topology)]] and we get an isomorphism. | ||
| + | |||
| + | I hope this is the similarity that is expected. | ||
<noinclude> | <noinclude> | ||
==Notes== | ==Notes== | ||
Latest revision as of 00:13, 12 October 2016
Section B
Question 8
Suppose that [ilmath]f:X\rightarrow Y[/ilmath] is a continuous function that is also surjective, and let [ilmath]\sim[/ilmath] denote the equivalence relation induced by [ilmath]f[/ilmath] on [ilmath]X[/ilmath].
Show that if [ilmath]X[/ilmath] is compact and [ilmath]Y[/ilmath] is Hausdorff then [ilmath]\frac{X}{\sim} [/ilmath] is homeomorphic to [ilmath]Y[/ilmath].
Which theorem of group theory does this resemble?
Proof
Recall the result of question 5:
Using the linked theorem we instantly obtain:
- [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] that is a continuous bijection
- We must show this is a homeomorphism.
Suppose [ilmath]X[/ilmath] is compact, then [ilmath]\frac{X}{\sim} [/ilmath] is compact also as the image of a compact set is compact and [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] is continuous (see quotient topology for more details)
We then apply the "compact-to-Hausdorff theorem" which shows us that [ilmath]\bar{f} [/ilmath] is actually a homeomorphism
This is similar to the first group isomorphism theorem (in the case the map is surjective) certainly in a categorical sense. The group theorem "factors through the kernel of the morphism" where as this "factors through the equivalence relation induced by the morphism" and they both yield an isomorphism (a homeomorphism is the term for a topological isomorphism. Here are diagrams:
- For any (surjective) group homomorphism [ilmath]\varphi:G\rightarrow H[/ilmath] (for arbitrary groups) we get a group isomorphism, [ilmath]\bar{\varphi} [/ilmath]. [ilmath]\xymatrix{ G \ar[r]^\varphi \ar[d]_\pi & H \\ {G/\text{Ker}(\varphi)} \ar@{.>}[ur]_{\bar{\varphi} } }[/ilmath], and,
- For any (surjective) continuous map, [ilmath]f:X\rightarrow Y[/ilmath] (for an arbitrary compact space [ilmath]X[/ilmath], and arbitrary Hausdorff space [ilmath]Y[/ilmath]) we get a homeomorphism/topological isomorphism [ilmath]\bar{f} [/ilmath]. [ilmath]\xymatrix{ X \ar[d]_\pi \ar[r]^f & Y \\ \frac{X}{x_1\sim x_2\iff f(x_1)=f(x_2)} \ar@{.>}[ur]_(.6){\bar{f} } }[/ilmath]
In both instances the "denominator" of the quotient depends on the map, factoring (as in passing to the quotient (function) is involved, rather than passing to the quotient (topology) and we get an isomorphism.
I hope this is the similarity that is expected.
Notes
References
