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(Note: in the post below, whenever I use the word "composition", I mean "defined composition", where the domain of the second morphism is the codomain of the first one; likewise, every composition expression "$g\;\;{\scriptstyle \circ}\;\;f\;$" implies the assertion dom($g$) = cod($f$).)

Arbib and Manes define an image factorization system for a category $\mathbf{K}$ as a pair $(\mathbf{E}, \mathbf{M})$ where $\mathbf{E}$ and $\mathbf{M}$ are classes of morphisms in $\mathbf{K}$ satisfying the following four axioms:

IFS1: $\mathbf{E}$ and $\mathbf{M}$ are both closed under compositions.

IFS2: The members of $\mathbf{E}$ are epic and the members of $\mathbf{M}$ are monic.

IFS3: All of $\mathbf{K}$'s isomorphisms belong to $\mathbf{E}\cap \mathbf{M}$.

IFS4: Every $f:A\to B$ in $\mathbf{K}$ can be factored as $f = m\;\;{\scriptstyle \circ}\;\;e$, where, $e\in \mathbf{E}$ and $m\in \mathbf{M}$, such that, for any other $e'\in \mathbf{E}$ and $m'\in \mathbf{M}$ with $f = m'\;\;{\scriptstyle \circ}\;\;e'$, there exists an isomorphism $\psi$ such that $e' = \psi\;\;{\scriptstyle \circ}\;\;e$ and $m' = m\;\;{\scriptstyle \circ}\;\;\psi^{-1}$.

Suppose that for some category $\mathbf{K}$ there exists an image factorization system $(\mathbf{E}, \mathbf{M})$. Can there be some other class $\mathbf{M}' \neq \mathbf{M}$ such that $(\mathbf{E}, \mathbf{M}')$ is also an image factorization for $\mathbf{K}$?

Thanks!

PS:

Above, I reworded Arbib-Manes' definition of image factorization system. Just in case my rewording is not as faithful to the original as I think it is, I copy the original passages verbatim below.

[p. 38]

13 DEFINITION: An image factorization system for a category $\mathbf{K}$ consists of a pair $(\mathbf{E}, \mathbf{M})$ where $\mathbf{E}$ and $\mathbf{M}$ are classes of morphisms in $\mathbf{K}$ satisfying the following four axioms:

IFS1: If $e:A\to B\in\mathbf{E}$ and $e':B\to C\in\mathbf{E}$ then $e'\cdot e:A\to C\in\mathbf{E}$. Dually, if $m:A\to B\in\mathbf{M}$ and $m':B\to C\in\mathbf{M}$ then $m'\cdot m:A\to C\in\mathbf{M}$.

IFS2: If $e:A\to B\in\mathbf{E}$, $e$ is an epimorphism. Dually, $m:A\to B\in\mathbf{M}$, $m$ is a monomorphism.

IFS3: If $f:A\to B$ is an isomorphism, then $f\in\mathbf{E}$ and $f\in\mathbf{M}$.

IFS4: Every $f:A\to B$ in $\mathbf{K}$ has an $\mathbf{E}$-$\mathbf{M}$ factorization which is unique up to isomorphism. More precisely, there exists an E-M factorization $(e, m)$ of $f$, meaning $e\in\mathbf{E}, m\in\mathbf{M}$ and $f = m\cdot e$, (so that there exists an object—call it $f(A)$—such that $e$ has the form $e:A\to f(A)$ and $m$ has the form $m:f(A)\to B$), and this factorization is unique in the sense that if $(e', m')$ is another such factorization—$f = m'\cdot e', e'\in\mathbf{E}, m'\in\mathbf{M}$—then there exists an isomorphism [$\psi: f(A)\to C=\mathrm{cod}(e')=\mathrm{dom}(m')$] . . . with $\psi\cdot e= e', m'\cdot\psi = m$.

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1 Answer 1

up vote 2 down vote accepted

The answer is yes. You can find it in the book "Abstract and Concrete Categories" (available online), Prop. 14.6. More specifically, $E$ and $M$ determine each other via the diagonalization property.

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You mean "The answer is no". :) kjo was asking if it's possible to have two factorizations systems with the same E and different M's. –  Omar Antolín-Camarena Mar 1 '12 at 17:22
    
@MartinBrandenburg: That's exactly what I was looking for. Thanks! –  kjo Mar 1 '12 at 17:31
    
@Omar: I refer to the question in the title. –  Martin Brandenburg Mar 1 '12 at 17:42
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