Is it true that any morphism in any category can be written as a combination of monomorphism and epimorphism? In SET and categories where monomorphism is an injective function and epimorphism is a surjective function is it true. But I am interested about the issue in any category. How to prove that statement or there is a counterexample?
2026-02-22 23:23:28.1771802608
Morphism=monomorphism•epimorphism?
674 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 MONOMORPHISMS
- Monomorphisms, unclear basic property, Functor
- Relation between homomorphisms and monomorphims of finite groups
- Is the unique subobject also mono?
- Epimorphism and monomorphism explained without math?
- Morphism=monomorphism•epimorphism?
- $\lambda$-pure morphisms in $\lambda$-accessible categories are monos, unclear proof
- Investigate whether the given transformation is a monomorphism / epimorphism. Find image and kernel
- Proof verification: An arrow which is monic under a faithful functor is itself monic
- Every $\lambda-$pure morphism in a locally $\lambda-$presentable category is a regular monomorphism
- Are all split epimorphisms effective?
Related Questions in EPIMORPHISMS
- If $f:G→ℤ_{8}$ be a group epimorphism , then what can be said about $G$?
- Is surjection an epimorphism without the axiom of choice?
- Epimorphism, in the category of commutative rings with unity , with domain a field is an isomorphism?
- Epimorphism and monomorphism explained without math?
- Is the inclusion of $\mathbb{Z}\to\mathbb{Q}$ an epimorphism?
- Strict Epimorphism of Schemes
- Fundamental homomorphism theorem (epimorphism)
- Why is epimorphism not defined as follows?
- Epimorphism and generator of Mod-$ R $
- Example of a NON-effective epimorphism
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?
In general there is no reason for this to be true. A simple counterexample is the category with one object $x$ and exactly one non-identity arrow $f:x\to x$, where $f\circ f=f$. In this category $f$ is neither a monomorphism nor an epimorphism, so the only monomorphism/epimorphism is the identity, and $f$ is not a composition of the identity with itself. More generally, if $\mathcal{C}$ is a category where every epimorphism or monomorphism is an isomorphism, then clearly an arrow that is not an isomorphism cannot be decomposed like that (though I can't think of any other example of this kind).
It is true, however, for most concrete categories that one usually considers; in particular, a category where every morphism factors as a strong epimorphism followed by a monomorphism and strong epimorphisms are stable under pullbacks is called a regular category, and any quasivariety of universal algebra, any abelian category or any quasitopos is regular. Note that the category of topological spaces is not regular, but it still has the property that every arrow factors as an epi followed by a mono (in fact its dual is regular).