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5

One way to formalize the idea is to say the following: the category of classical propositional theories is a reflective subcategory of the category of intuitionistic propositional theories. In order to make sense of that claim, it's best to pass from "theories" to "algebras." Each classical theory corresponds to a Boolean algebra, and each intuitionistic ...


5

If $S$ has a categorification $\mathrm{Decat}(\mathcal{C}) \to S$ and $T$ has a categorification $\mathrm{Decat}(\mathcal{D}) \to T$, then a categorificiation of a map $S \to T$ is a functor $\mathcal{C} \to \mathcal{D}$ such that the induced map $\mathrm{Decat}(\mathcal{C}) \to \mathrm{Decat}(\mathcal{D})$ makes the diagram $$\begin{array}{cc} ...


4

I've seen the notation $\mathsf{CAlg}_R$ in many places. I prefer $\mathsf{CAlg}(R)$ in order to stress the functoriality. This is also used in Yves Diers' work. Many papers restrict to commutative rings and algebras in the first place and therefore just write $\mathsf{Alg}_R$ (which might be confusing - but this is just a local notation). More generally, ...


4

Check out the new book (amazon-link) Tom Leinster, Basic Category Theory, Cambridge Studies in Advanced Mathematics, Vol. 143, 2014


3

The polynomial ring $A [x]$ is the coproduct of $A$ and $\mathbb{Z} [x]$. One way to see this is to note that the category of (commutative unital) $A$-algebras is isomorphic to the slice category $^{A /} \mathbf{CRing}$, and the forgetful functor $^{A /} \mathbf{CRing} \to \mathbf{CRing}$ has a left adjoint, namely $B \mapsto A \otimes_{\mathbb{Z}} B$. ...


3

The trick is to embed the category of abstract simplicial complexes inside the category of symmetric simplicial sets (= functor $\mathbf{F}^\mathrm{op} \to \mathbf{Set}$, where $\mathbf{F}$ is the category of positive finite cardinals): this can be done by sending an abstract simplicial complex $X$ to the symmetric simplicial set ...


3

Notice that $V$ and $V'$ are orthogonal with respect to $b \oplus b'$, i.e. we have $(b \oplus b')(v \oplus v')=0$ and $(b \oplus b')(v' \oplus v)=0$ for all $v \in V$, $v' \in V'$. In fact, $(V \oplus V',b \oplus b')$ represents the subfunctor of $\hom((V,b),-) \times \hom((V',b'),-)$ which consists of those pairs of morphisms $f : (V,b) \to (W,c)$, $g : ...


2

Assuming we're working with small categories $\mathcal{C,D}$, the problem with images not being categories requires, for a functor $F$, the map of sets $F: Ob(\mathcal{C}) \to Ob(\mathcal{D})$ not being injective (though this isn't sufficient to insure a problem). The issue that arises is that there may be objects and morphisms $f:W \to X, g:Y \to Z$ in ...


2

The source of the problem, as described by JHance, can be realized as the smallest 'free' example of a functor $F\colon C\to D$ whose image is not a category. Let $C$ be the category with four (distinct) objects $w$, $x$, $y$, $z$ and the two arrows $w\to x$ and $y\to z$ (besides the identities), and let $D$ be the category with the objects $0$, $1$, $2$ and ...


1

I have no idea what kind of generalizations are you looking for, but the above statement is completely false even in case of ordinary categories (and you can easily find a counterexample). It is true, however, under some additional assumptions (like, finite completeness). Check the following question: ...


1

Well, sets and maps. In set theory, a map from $A$ to $B$ is a certain subset of $A \times B$. If the map factors through some subset $B'$ of $B$, we get two maps $A \to B$ and $A \to B'$ which are equal as sets. For example, for a set theorist, $\emptyset$ is a map from $\emptyset$ to any set. But for a category theorist, a morphism has to have a specified ...


1

You can define a "simplex" as having an orientation, thus getting an easier answer. A $k$-simplex is the convex hull of a set of $k+1$ points. But what does it mean for a set to have $k+1$ points? That there is a bijection from $\{1,2,\dots,k+1\}$. So simply define "$k$-simplex" in terms of a map $\{1,2,\dots,k+1\}\to\mathbb R^n$ and you can pick an ...


1

Just don't call it a pullback. Yes this is ok.


1

Your idea is absolutely right. The map $\varphi_{(S,Q)}$ maps a morphism $\psi \colon C(S) \rightarrow Q$ (in $\mathrm{CompInfDist}_\vee$) to the morphism $\psi \circ \iota_S$ (in $\mathrm{InvSem}$), where $\iota_S \colon S \hookrightarrow C(S)$ is the map you denoted as $\iota$.


1

Introduction to Category Theory by Harold Simmons is a nice and gentle way to get into category theory with plenty of exercises (and full solutions!). I'm an undergrad as well, and I worked through this book before moving on to Categories for the Working Mathematician because it is more leisurely. More to the point of your question, Intro to Category Theory ...


1

Let $n \in \mathbb{N}_{\geq 1}$. The simplicial category $\mathfrak{C} [\partial \Delta^n]$ has as objects the set $\{0,\ldots,n\}$, and the simplicial set $Hom_{\mathfrak{C}[ \partial \Delta^n]}(i,j)$, for $0 \leq i < j \leq n$ and $(i,j) \neq (0,n)$, is $N(P_{i,j})$, the nerve of the partially ordered set of subsets $I \subseteq \{ i,\ldots,j \}$ with ...


1

Let's do part of the task, to get you the feeling how to proceed with the other parts: $\emptyset$ is initial in the category of sets (and maps): If $A$ is any set, we need to exhibit a map $f\colon \emptyset\to A$ and show that it is unique. To produce a map we need to specify an element $f(x)$ for each $x\in\emptyset$. But there is nothing to be done! ...



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