# Tag Info

6

If you want an honest functor, you can simply choose any representative of the isomorphism class in question; it doesn't have to be "canonical". That is, for every pair $(X,Y)$ of $Z$-schemes, you can choose some triple $(P_{X,Y},p_{X,Y},q_{X,Y})$ where $P_{X,Y}$ is a $Z$-scheme and $p_{X,Y}:P_{X,Y}\to X$ and $q_{X,Y}:P_{X,Y}\to Y$ are maps of $Z$-schemes ...

6

The function you asked is related to the function $$G(x) = \sum_{n=0}^{\infty} \frac{x^{n^2}}{(1-x)\cdots(1-x^n)}$$ appearing in Rogers-Ramanujan identity by $$\sum_{n=0}^{\infty} \Bigg( \prod\limits_{i=0}^{n-1} (q^n-q^i) \Bigg)^{-1} = G(1/q).$$ I learned that this has been studied in combinatorics since the coefficients in the series expansion admit ...

4

This is Theorem 6.12 of Paul Mitchener's article $C^\ast$-categories.

4

This is not even possible on the level of objects, let alone morphisms; not every group is a commutator subgroup.

4

The tensor product is defined up to canonical isomorphism, which is different from being defined up to isomorphism: there is a universal map $M\times N \to M\otimes N$ through which any bilinear map $M \times N \to R$ factors, and if you give one construction of the tensor product (e.g. generators and relations), and I give another one (e.g.perhaps I ...

3

To summarize the discussion from the comments: yes, every example of a thing you want to call a universal property (for an object in a category) has this form, even if it sometimes takes some thinking to cook up the relevant functor. (But I promise that this is not work relative to what work looks like in the rest of mathematics.) I prefer to think of the ...

3

Every object has at least one arrow (the identity). So there is only one category with 1 arrow: one object and the identity. For two arrows: we either have 2 objects (and no more) that both only have the identity (that's 1), or one object $X$ with identity $1_X$ and another arrow $f$. But then either $f \circ f = f$ or $f \circ f = 1_X$ (it has to be one of ...

3

Yes, this is true (and more generally, the same construction works in pretty much any "concrete category"). I should mention that I have never seen the term "relative product" used to refer to this construction; more common terms are "pullback" or "fiber product". As a deeper explanation for why relative products look the same (while other constructions ...

3

A functor does several things at once: It sends objects in $\def\A{\mathbf A}\A$ to objects in $\def\B{\mathbf B}\B$. If this map $F\colon \def\O{\mathop{\rm Obj}}\O\A \to \O \B$, $A \mapsto FA$ is injective, one calls $F$ injective on objects. It sends morphims in $\A$ to morphisms in $\B$. If this map $F \colon \def\M{\mathop{\rm Mor}}\M\A \to \M\B$, ...

3

Isomorphisms are absolutely essential to category theory, and in particular the idea that isomorphic objects are "the same" is perhaps the single most important concept in all of category theory. As you observe, you can define isomorphisms without requiring the existence of identities. However, this definition is still a bit problematic in a few ways. ...

3

"Natural" can be given a precise meaning using the concept of natural transformations. For example, loosely speaking, for any group $G$ there is a "natural" homomorphism $[G, G] \to G$, where $[G, G]$ is the commutator subgroup of $G$, and similarly there is a "natural" homomorphism $G \to G/[G, G]$. In fact there is a "natural" short exact sequence $$1 \to ... 2 When considering the 'one object case', this is basically the question of semigroups vs. monoids (or, in additive fashion, rings with or without units). And such a debate can go into philosophical deepness, where both sides have their own truth. (Note that non unital categories do exist.) So.. why to use identity elements? Why not? They can be freely ... 2 Let \mathcal{S} be a category with finite products and let R be a ring in \mathcal{S}. There is an \mathcal{S}-enriched Lawvere theory \mathcal{T}_R where \mathcal{T}_R (n, m) = R^{m \times n} with composition defined by matrix multiplcation. We can define models of \mathcal{T}_R in \mathcal{S} to be an object M together with morphisms ... 2 In a preordered set (X,\le), a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of X. This shows that \omega, which lacks a maximum element, is neither complete nor cocomplete as a category. This is due to the ... 2 So to clarify, this is true in a spherical category (when you assume that left and right traces are equal, this is the definition of a spherical category). The best way (IMO) to prove this is through graphical calculus. Here is a picture (which I hope is not too horrible). The only non-obvious equality is the first: it follows from the monoidality of the ... 2 You aren't just proving that \mathcal C is complete, you know that it's complete in a specific way, ie. that limits are calculated the same way both in \mathcal C and in \mathcal D, or in other words, that the canonical map η : \lim X → r \lim X must be an isomorphism. (This is because the inclusion, being by assumption right adjoint, must preserve ... 2 Let F(-) = \mathcal J(k,-) and let's go back to your direct calculation of the colimit in \mathsf{Set}. The colimit of F \colon \mathcal J \to \mathsf{Set} is given by$$ \operatorname{colim}F \simeq \coprod_{j\in \mathcal J}F(j) \big/ {\sim} $$where \sim identifies x \in F(j) with F(f)(x) \in F(j') for any f\colon j\to j' in \mathcal J. ... 2 Let F (-) = \mathcal{J} (k, -). Then:$$\mathbf{Set} (\varinjlim\nolimits_\mathcal{J} F, X) \cong [\mathcal{J}, \mathbf{Set}] (F, \Delta X) \cong X \cong \mathbf{Set} (1, X)$$Hence, \varinjlim\nolimits_\mathcal{J} F \cong 1. 2 There are tons and tons of such functors. Given any such functor you can compose it with either an endofunctor of \text{Set} or an endofunctor of \text{Mon}, and there are tons of those. So let me first restrict attention to the nicest ones, which are the ones that preserve colimits. \text{Set} turns out to be the free cocomplete category on a point: ... 2 Yes this is true. Let f:Y\rightarrow X be any morphism in C. Then form the commutative square$$ \require{AMScd}\begin{CD}Y@>{f}>>X \\ @V{f}VV@|\\X@=X\end{CD}$$If all square is a pullback, then f is a pullback of an isomorphism, hence an isomorphism. 1 The answer depends on definitions of category and functors, but the idea is that functor F\colon \mathcal C\to\mathcal D acts on objects and morphisms: for every object c in category \mathcal C, Fc is object in category \mathcal D for every two objects c and c' in \mathcal C and every morphism f\in \operatorname{Hom}_{\mathcal C}(c,c'), ... 1 Such functors are precisely the pseudomonomorphisms in the 2-category of categories. (A pseudomonomorphism is a morphism f : X \to Y in a bicategory such that$$\require{AMScd} \begin{CD} X @= X \\ @| @VV{f}V \\ X @>>{f}> Y \end{CD} is a bicategorical pullback square.)

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Second question first: the coproduct of morphisms is the same thing as the pushout. You're right that the coproduct of objects is the direct sum. Your discussion of the fiber product is quite confused: it's not actually clear to me what you're trying to show. In a literal sense, it does not require proof that the fiber product exists, at least, if you ...

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This answer is just a straw man that meets the first part of your question about no finite sets, but not the second part about shifting cardinalities: take the subcategory $\cal U$ of $\mathsf{Set}$, comprising all objects of the form $\Bbb{N} \times X$ and all morphisms from $\Bbb{N} \times X \to \Bbb{N} \times Y$ of the form $(i, x) \mapsto (i, f(x))$ ...

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First as clarification: If $f$ is a morphism we say $id_{d(f)}$ is the identity morphism on the domain of $f$ and $id_{c(f)}$ is the identity on the codomain of $f$. lemma: Let be $C$ a category with epic-equalizer factorization. Let $f$ be a morphism and $e_f$, $i_f$ its epic-equalizer factorization. $f$ is equalizer if and only if $e_f$ is an ...

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