[0MKE]
Lemma 13.8. One has .
[0MKF]
Proof. By Lemma 13.7, it is enough to check that the generators of and of are contained in . For that assume that is a generator of and let be an inclusion. We wish to demonstrate that for all we have that
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is in .
Recall that
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There are several cases:
- (a)
. In this trivial case (13.8.1) reduces to .
- (b)
The morphism factors as . In this case the fiber product is either or empty. In the latter case (13.8.1) is an isomorphism, and in the former case it is . Notice that this case covers .
- (c)
and the map does not factor through . In this case it follows that is not in , and hence
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is representable with . Moreover the map
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consists of a surjective map (which specifies a unique ; the inverse image of consists of all elements strictly less than ) together with map . A direct calculation shows that in this situation (13.8.1) is either an isomorphism or a generator of .
- (d)
, the map
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is a suspension, and the map does not factor through . It follows that is the suspension of a map. This case then follows by induction and Lemma 13.3.
- (e)
The final case is when , the map is not a suspension, nor in , and the map does not factor through . In this case (13.8.1) is a map of the form described in Lemma 13.5. ∎