View source for Exercises:Saul - Algebraic Topology - 5/Exercise 5.6

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==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}} (with the [[continuous map]] of the retraction being {{M|r:X\rightarrow A}}). Lastly take i {{M|i:A\rightarrow X}} to be the [[inclusion map]], {{M|i:a\mapsto 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 as a [[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!
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==Notes==
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==References==
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