351059c72074 — Benedikt Fluhr <http://bfluhr.com> 7 years ago
Some Parts rewritten
3 files changed, 52 insertions(+), 46 deletions(-)

M 00_10_someEquivalences.md
M 00_11_negEnhJoinTrees.md
M 00_12_EqualityOfInterlDist.md
M 00_10_someEquivalences.md +1 -1
@@ 102,7 102,7 @@ is a $1$-homomorphism on this $D$-catego

Now let $f \colon X \rightarrow \R$ a continuous function.

-* **Lemma.**
+(@etaCEIso) **Lemma.**
The homomorphism $(\eta^{\pi^2} \circ \mathcal{C} \circ \mathcal{E})_f$
from $\mathcal{C} \mathcal{E} f$ to
$\pi^2_p \pi^2_* \mathcal{C} \mathcal{E} f$

M 00_11_negEnhJoinTrees.md +8 -5
@@ 13,14 13,17 @@ but we will need it later and we don't i
more notation.)
Now we define the smoothing functor $\mathcal{S}$ on $\mathbf{D}$.
The easiest part of the definition are the endofunctors.
-Moreover this endofunctors exist on the whole category
-of $\R$-spaces and not just $\mathbf{D}$.
-Now let $f \colon X \rightarrow \R$ be a continuous function,
+Moreover these endofunctors exist on the whole category
+of $(-\infty, \infty]$-spaces and not just $\mathbf{D}$.
+Now let $f \colon X \rightarrow (-\infty, \infty]$ be a continuous function,
then we set
$\mathcal{S}((a, b))(f) := f - b$ for all $(a, b) \in -D$.
-And for any homomorphism $\varphi$ in in the category of $\R$-spaces we set
+And for any homomorphism $\varphi$
+in the category of $(-\infty, \infty]$-spaces we set
$\mathcal{S}((a, b))(\varphi) := \varphi$.
-And this defines the endofunctors an all $\R$-spaces.
+And this defines the endofunctors
+$\{\mathcal{S}(\mathbf{a}) ~|~ \mathbf{a} \in -D\}$ on all
+$(-\infty, \infty]$-spaces.
Now let $f \colon X \rightarrow \R$
be an $\R$-space,
let $r \in \R$, and let $(a, b), (c, d) \in -D$

M 00_12_EqualityOfInterlDist.md +43 -40
@@ 6,12 6,46 @@ We aim to show that the Reeb precosheaf
defines a $1$-homomorphism from $\mathbf{D}$ to $\mathbf{C}$.
Now in order for this statement even to make any sense,
the image of $\mathbf{D}$ under $\mathcal{C}$ should lie in $\mathbf{C}$.
-The endomorphisms of the smoothing functor $\mathcal{S}'$ on $\mathbf{C}$
-were defined for all precosheaves on $\overline{D}$ however.
-So it still makes sense, even before we know whether
-the image of $\mathcal{C}$ lies in $\mathbf{C}$,
-to talk about the compatibility of $\mathcal{C}$ with the endomorphisms
-associated by the smoothing functors.
+(@imgC) **Lemma.**
+The image of $\mathbf{D}$ under $\mathcal{C}$
+is part of the subcategory $\mathbf{C}$.
+Now let $f \colon X \rightarrow \R$ be a bounded constructible
+$\R$-space.
+To proof the previous lemma it suffices to show that
+$\mathcal{C} (r + \mathcal{E} f)$ and
+$\mathcal{C} (r + \mathcal{R} \mathcal{E} f)$
+are an object of $\mathbf{C}$ for all $r \in (-\infty, \infty]$.
+We consider the case $r = 0$ first.
+By lemma @etaCEIso the precosheaf $\mathcal{C} \mathcal{E} f$
+is an object of $\mathbf{C}$.
+By lemma @lemEpigraphConstructible from
+[the last appendix](#skeleton-epigraph)
+the projection $\mathcal{E} f$ from $\epi f$ to $\overline{\R}$
+is a constructible $\overline{\R}$-space.
+This in conjunction with lemma @CpiIso
+yields the following
+(@CpiEIso) **Lemma.**
+The homomorphism $(\mathcal{C} \circ \pi \circ \mathcal{E})_f$
+is an isomorphism of precosheaves.
+Now $\mathbf{C}$ is closed under isomorphisms and thus also
+$\mathcal{C} \mathcal{R} \mathcal{E} f$ lies in $\mathbf{C}$.
+Now we consider the case of $r$ not necessarily being $0$.
+We note that for any continuous function
+$g \colon Y \rightarrow (-\infty, \infty]$ and
+$r \in (-\infty, \infty]$ we have
+$r + g = \mathcal{S}((-r, -r)) (g)$
+by definition of $\mathcal{S}$.
+We now recall that we also defined the endofunctors
+associated to the smoothing functor
+$\mathcal{S}'$ for $\mathbf{D}$
+on the whole category of set-valued precosheaves on $\overline{D}$
+and not just $\mathbf{D}$.
+With $\mathbf{D}$ being invariant under these endofunctors,
+lemma @imgC follows from the following

* **Lemma.**
For all $(a, b) \in -D$ we have

@@ 25,41 59,10 @@ Then we have
$\Delta \circ \mathcal{S}((a, b))(f) = S^{(-b, -b)} \circ \Delta \circ f$
and this implies the claim.

-Now in order to show that the image of $\mathbf{D}$ under
-$\mathcal{C}$ lies in $\mathbf{C}$,
-we just have to show that the image of each epigraph and each join tree
-of a bounded constructible $\R$-space is in $\mathbf{C}$.
-Because everything else in $\mathbf{D}$ is just a shift,
-which is compatible with the smoothing functors' endomorphisms by the above,
-and because $\mathbf{D}$ is invariant under these endomorphisms.
-(@CpiEIso) **Lemma.**
-Let $f \colon X \rightarrow \R$ be a continuous function,
-then $(\eta^{\pi^2} \circ \mathcal{C} \circ \mathcal{E})_f$
-is an isomorphism of precosheaves.
-By this lemma $\mathcal{C} \mathcal{E} f$ lies in $\mathbf{C}$
-for any continuous function $f$.
-Now suppose $f \colon X \rightarrow \R$ is a bounded constructible
-$\R$-space, then the epigraph of $f$ is also constructible
-by lemma @lemEpigraphConstructible from
-[the last appendix](#skeleton-epigraph).
-This in conjunction with lemma @CpiIso
-yields the follwoing
-(@2ReebIso) **Lemma.**
-The homomorphism $(\mathcal{C} \circ \pi \circ \mathcal{E})_f$
-is an isomorphism of precosheaves.
-Now $\mathbf{C}$ is closed under isomorphisms and thus also
-$\mathcal{C} \mathcal{R} \mathcal{E} f$ lies in $\mathbf{C}$.
-This in conjunction with the above observations implies
-that the image of $\mathbf{D}$ under $\mathcal{C}$ lies in
-$\mathbf{C}$.
Next we show that $\mathcal{C}$ is compatible with the natural
transformations of the smoothing functors.
-To this end let $\mathcal{a}, \mathcal{b} \in -D$ with
-$\mathcal{a} \preceq \mathcal{b}$.
+To this end let $\mathbf{a}, \mathbf{b} \in -D$ with
+$\mathbf{a} \preceq \mathbf{b}$.

* **Lemma.**
We have

@@ 111,7 114,7 @@ From this and the previous corollary we
\mathcal{S}'(\mathbf{b}) \mathcal{C} \mathcal{R} \mathcal{E} f .
}

-Now by lemma @2ReebIso the vertical arrows are isomorphisms
+Now by lemma @CpiEIso the vertical arrows are isomorphisms
and thus the naturality of $\mathcal{S}'(\mathbf{a} \preceq \mathbf{b})$
implies that
\$(\mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b})