f3c974521ba0 — Benedikt Fluhr <http://bfluhr.com> 5 years ago
Change of Terminology: D-category instead of D-module

We use the term D-category where we previously used the term D-module
to avoid a clash of terminology,
since D-modules already exist and are something completely different.
This has been pointed out by Michael Catanzaro.
1 files changed, 7 insertions(+), 7 deletions(-)

M poster.tex

M poster.tex +7 -7
@@ 198,15 198,15 @@
for all $$(a, b; c, d) \in \mathcal{D}$$.
}
\column{0.5}
-  \block{Interleavings in $$D$$-modules}{
-    \begin{definition}[$$D$$-modules]
-      A \emph{$$D$$-module} is a category $$\mathcal{C}$$
+  \block{Interleavings in $$D$$-categories}{
+    \begin{definition}[$$D$$-categories]
+      A \emph{$$D$$-category} is a category $$\mathcal{C}$$
with a strict monoidal functor $$\mathcal{S}$$ from $$D$$
to the category of endofunctors on $$\mathcal{C}$$.
We refer to $$\mathcal{S}$$ as the
\emph{smoothing functor of $$\mathcal{C}$$}.
\end{definition}
-    Now let $$\mathcal{C}$$ be a $$D$$-module
+    Now let $$\mathcal{C}$$ be a $$D$$-category
with smoothing functor $$\mathcal{S}$$.
For $$a \leq 0$$, $$b \geq 0$$, and an object $$A$$ in $$\mathcal{C}$$
we get two things,

@@ 273,10 273,10 @@
and $$\mu(A, B)$$ the
\emph{relative interleaving distance}.
\end{definition}
-    Now let $$\mathcal{C}'$$ be another $$D$$-module with smoothing
+    Now let $$\mathcal{C}'$$ be another $$D$$-category with smoothing
functor $$\mathcal{S}'$$.
\begin{definition}
-      A $$1$$-homomorphism of $$D$$-modules
+      A $$1$$-homomorphism of $$D$$-categories
from $$\mathcal{C}$$ to $$\mathcal{C}'$$
is a functor from $$\mathcal{C}$$ to $$\mathcal{C}'$$ such that
\[

@@ 337,7 337,7 @@
to a category $$\mathcal{C}$$.
\begin{definition}
A \emph{persistence-enhancement of $$F$$}
-      is the structure of a $$D$$-module on $$\mathcal{C}$$
+      is the structure of a $$D$$-category on $$\mathcal{C}$$
together with a $$1$$-homomorphism $$\tilde{F}$$ from
$$\mathcal{F}$$ to $$\mathcal{C}$$ such that
$$\tilde{F}((\_, \mathbf{o})) = F$$.