The title says it all really. Given an additive category $\mathcal{A}$, is having all kernels and arbitrary products sufficient to conclude that it has all limits? Dually, is having all cokernels and arbitrary coproducts sufficient to conclude it has all colimits? I feel like this should be really quite simple, but I'm having trouble finding a reference that actually states this.
2026-03-25 03:20:17.1774408817
Constructing limits in an additive category given the existence of products and kernels
193 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
There are 1 best solutions below
Related Questions in CATEGORY-THEORY
- (From Awodey)$\sf C \cong D$ be equivalent categories then $\sf C$ has binary products if and only if $\sf D$ does.
- Continuous functor for a Grothendieck topology
- Showing that initial object is also terminal in preadditive category
- Is $ X \to \mathrm{CH}^i (X) $ covariant or contravariant?
- What concept does a natural transformation between two functors between two monoids viewed as categories correspond to?
- Please explain Mac Lane notation on page 48
- How do you prove that category of representations of $G_m$ is equivalent to the category of finite dimensional graded vector spaces?
- Terminal object for Prin(X,G) (principal $G$-bundles)
- Show that a functor which preserves colimits has a right adjoint
- Show that a certain functor preserves colimits and finite limits by verifying it on the stalks of sheaves
Related Questions in ABELIAN-CATEGORIES
- What is the monomorphism that forms the homology group?
- Injective objects in a category
- Category of complexes
- Snake lemma and regular epi mono factorization
- A question to Weibel’s IHA lemma 2.6.14 Part 2
- Why do the finitely generated subsheaves of a sheaf form a directed system?
- Supremum of a family of subobjects in an abelian category
- Opposite effective classes in a Grothendieck group
- Question about $\mbox{Ext}$ groups in abelian categories
- How to show that $\mathsf{Ab}$(Category of Abelian Groups) is an abelian category?
Related Questions in LIMITS-COLIMITS
- Show that a certain functor preserves colimits and finite limits by verifying it on the stalks of sheaves
- Prove that a "tensor product" principal $G$-bundle coincides with a "pullback" via topos morphism
- What prevents me from starting the cone in the objects mapped by the diagram
- Two morphisms $f, g : M \to L$ are equal as long as they are equal under the limit $L$.
- Separating coproducts by its index set
- Preservation of coproducts by coproducts
- Categorification of commutative sum
- Colimit of a directed system of modules
- Pullback square with two identical sides
- Subring of an inverse limit is the entire thing if maps to pieces are surjective
Related Questions in ADDITIVE-CATEGORIES
- Is the kernel of projections in an additive category preserved by additive functors?
- Mapping cone and additive functors
- Additive Yoneda Lemma
- In an additive category, why a morphism between complexes can be factored as a composition of a homotopy equivalence and a monomorphism?
- Proving the direct sum of two distinguished triangles is a distinguished triangle
- Interpretation of preadditive categories as rings.
- Submodules in a preadditive category with one object.
- Identity making each $\operatorname{Hom}_\mathcal{A}(A,B)$ an abelian group in an additive category the Zero map?
- axioms of additive categories
- How to think about the octohedral axiom for triangulated categories?
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Find $E[XY|Y+Z=1 ]$
- Refuting the Anti-Cantor Cranks
- What are imaginary numbers?
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- How do we know that the number $1$ is not equal to the number $-1$?
- What are the Implications of having VΩ as a model for a theory?
- Defining a Galois Field based on primitive element versus polynomial?
- Can't find the relationship between two columns of numbers. Please Help
- Is computer science a branch of mathematics?
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- Generator of inertia group in function field extension
Popular # Hahtags
second-order-logic
numerical-methods
puzzle
logic
probability
number-theory
winding-number
real-analysis
integration
calculus
complex-analysis
sequences-and-series
proof-writing
set-theory
functions
homotopy-theory
elementary-number-theory
ordinary-differential-equations
circles
derivatives
game-theory
definite-integrals
elementary-set-theory
limits
multivariable-calculus
geometry
algebraic-number-theory
proof-verification
partial-derivative
algebra-precalculus
Popular Questions
- What is the integral of 1/x?
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- Is a matrix multiplied with its transpose something special?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- How to tell if a set of vectors spans a space?
- Calculus question taking derivative to find horizontal tangent line
- How to determine if a function is one-to-one?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- Is this Batman equation for real?
- How to find perpendicular vector to another vector?
- How to find mean and median from histogram
- How many sides does a circle have?
Yes, it is enough. Indeed, it is well-known (see for example the nLab, or Categories for the Working Mathematician, Chapter V, Section 2, Theorem 1) that a category has all limits if and only if it has arbitrary products and equalizers, so it is enough to prove that an additive category with kernels has equalizers. But given two maps $f,g:X\to Y$, for any map $h:Z\to X$ we have $fh=gh\Leftrightarrow (f-g)h=0$; thus the equalizer of $f$ and $g$ is the same thing as the kernel of $(f-g):X\to Y$.