Difference between revisions of "Exercises:Saul - Algebraic Topology - 5/Exercise 5.6"
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(Created page with "<noinclude> ==Exercises== ===Exercise 5.6=== </noinclude>Let {{Top.|X|J}} be a topological space and let {{M|A\in\mathcal{P}(X)}} be a {{link|retract|topology}} of {{M|X}}...") |
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Show that: {{M|H_*^s(X)\cong H_*^s(A)\oplus H_*^s(X,A)}} | Show that: {{M|H_*^s(X)\cong H_*^s(A)\oplus H_*^s(X,A)}} | ||
| − | ==== | + | ====Possible solution==== |
| + | I have proved: | ||
| + | * If {{M|f:X\rightarrow Y}} is a [[homotopy equivalence]] then {{M|f_*:H_n(X)\rightarrow H_n(Y)}} is a [[group isomorphism]] for each {{M|n\in\mathbb{N}_{\ge 0} }} | ||
| + | Then the [[corollary to the above]] | ||
| + | * If {{M|A\in\mathcal{P}(X)}} is a {{link|retract|topology}} of {{M|X}} (and {{M|r:X\rightarrow A}} is the [[continuous map]] of the retraction) then {{M|i_*:H_*(A)\rightarrow H_*(X)}} is a [[monomorphism]] ([[injection]]) ''onto'' (as in [[surjective]]) a ''{{link|direct summand|group theory}}'' of {{M|H_*(X)}} | ||
| + | ** If {{M|A}} is a [[deformation retraction]] of {{M|X}} ({{Caveat|presumably strong?}}) then {{M|i_*}} is an isomorphism. | ||
| + | * {{M|i:A\rightarrow X}} is the inclusion mapping. | ||
| + | |||
| + | To prove this corollary I show: | ||
| + | * {{M|H_*(X)\cong G\oplus H}} where: | ||
| + | ** {{M|G:\eq\text{Im}(i_*)}} and {{M|H:\eq\text{Ker}(r_*)}} | ||
| + | |||
| + | The second part of the statement (the deformation retraction part) is an immediate result of the first theorem, the second bit is proved without reference to it. So I should be good! | ||
<noinclude> | <noinclude> | ||
==Notes== | ==Notes== | ||
Revision as of 20:12, 14 February 2017
Exercises
Exercise 5.6
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be a retract of [ilmath]X[/ilmath] (with the continuous map of the retraction being [ilmath]r:X\rightarrow A[/ilmath]). Lastly take i [ilmath]i:A\rightarrow X[/ilmath] to be the inclusion map, [ilmath]i:a\mapsto a[/ilmath].
Show that: [ilmath]H_*^s(X)\cong H_*^s(A)\oplus H_*^s(X,A)[/ilmath]
Possible solution
I have proved:
- If [ilmath]f:X\rightarrow Y[/ilmath] is a homotopy equivalence then [ilmath]f_*:H_n(X)\rightarrow H_n(Y)[/ilmath] is a group isomorphism for each [ilmath]n\in\mathbb{N}_{\ge 0} [/ilmath]
Then the corollary to the above
- If [ilmath]A\in\mathcal{P}(X)[/ilmath] is a retract of [ilmath]X[/ilmath] (and [ilmath]r:X\rightarrow A[/ilmath] is the continuous map of the retraction) then [ilmath]i_*:H_*(A)\rightarrow H_*(X)[/ilmath] is a monomorphism (injection) onto (as in surjective) a direct summand of [ilmath]H_*(X)[/ilmath]
- If [ilmath]A[/ilmath] is a deformation retraction of [ilmath]X[/ilmath] (Caveat:presumably strong?) then [ilmath]i_*[/ilmath] is an isomorphism.
- [ilmath]i:A\rightarrow X[/ilmath] is the inclusion mapping.
To prove this corollary I show:
- [ilmath]H_*(X)\cong G\oplus H[/ilmath] where:
- [ilmath]G:\eq\text{Im}(i_*)[/ilmath] and [ilmath]H:\eq\text{Ker}(r_*)[/ilmath]
The second part of the statement (the deformation retraction part) is an immediate result of the first theorem, the second bit is proved without reference to it. So I should be good!
Notes
References
