0PBA
Lemma 7.4.19. Given , we have
and there is an injective morphism of groups
.
Let be a subset of that embeds in its
projection on . Define by
if and
otherwise. The morphism
induces an isomorphism of groups
.
We have if and only if .
Let be a map in . We have
,
,
and
.
0PBB
Proof. Let and let .
We have
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and
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In particular,
and
.
This shows that is injective. This shows also that given
, we have
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This shows the first equality and this shows that induces an injective
morphism of groups .
Taking quotients, we obtain an injective morphism of groups
compatible with
the order and with image .
Consider .
Given , we have
,
hence .
We have
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and
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Since
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it follows that .
Consider with . We have
for some minimal paths in
by Lemma 7.3.22.
Lemma 6.2.3 shows that
.
Lemma 6.2.2 shows now that .
∎