Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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 K as a pair (E, M) where E and M are classes of morphisms in K satisfying the following four axioms:

IFS1: E and M are both closed under compositions.

IFS2: The members of E are epic and the members of M are monic.

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

IFS4: Every $f:A\to B$ in 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 $m' = m\;\;{\scriptstyle \circ}\;\;\psi^{-1}$.

Then, in a subsequent exercise, they ask for a proof that, in any category K "in which every morphism can be factored as a coequalizer followed by a monomorphism, ... E = coequalizers and M = monomorphisms yields an image factorization system in K".

I can prove that such (E, M) satisfies IFS2 and IFS3, and that M satisfies IFS1, but I can't prove that for any $e_1, e_2 \in \mathbf{E}$, then also $e_2\;\;{\scriptstyle \circ}\;\;e_1 \in \mathbf{E}$ (i.e. IFS1 for E), nor have I managed much progress proving that all $m\;\;{\scriptstyle \circ}\;\;e$ factorizations ($e\in \mathbf{E}, m\in \mathbf{M}$) satisfy the uniqueness-up-to-isomorphism property of IFS4.

Any help with this proof (or a pointer to a book/paper that gives it) would be appreciated.



In my original post, I reworded Arbib-Manes' definition of image factorization system and the statement of the exercise in question. 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 K consists of a pair (E, M) where E and M are classes of morphisms in 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 K has an E-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 $\psi\cdot e= e', m'\cdot\psi = m$.

. . .

[p. 40]


. . .

9 Let K be a category in which every morphism factors as a coequalizer followed by a monomorphism. Prove that E = coequalizers, M = monomorphisms yields an image factorization system in K.

share|cite|improve this question
I don't think it's true that $\mathbf{E}$ (known in modern literature as "regular epimorphisms") is closed under composition. Further hypotheses on the category are needed. – Zhen Lin Jan 27 '12 at 21:11
Adding onto @Zhen Lin: We do have further hypotheses, though, namely that we always have factorizations. The nLab page for regular epimorphism says that $E\cap M$ contains only isomorphisms, and so it is enough to show that if $e\circ e'=e''=m\circ e''$, then we must have $m\in E$. It is easy to show that $m$ must be epi, I haven't tried to show that it is regular epi, though. – Aaron Jan 27 '12 at 21:36
@ZhenLin and Aaron: I added a verbatim quotation of the original text in Arbib-Manes, in case my rewording of it unintentionally introduced a meaning different from the one intended by the authors. – kjo Jan 27 '12 at 23:25
@Aaron: the info from nLab was useful, thanks. Proving that E $\cap$ M contains only isomorphisms somehow clarified the problem for me, even though this result is not used in the proof (in fact it follows readily from IFS4: if $\psi:A\to B$ is in E $\cap$ M, invoke IFS4 with $\psi = \psi\;{\scriptstyle \circ}\;1_A = 1_B\;{\scriptstyle \circ}\;\psi$). – kjo Jan 28 '12 at 5:46
up vote 2 down vote accepted

I managed to finish the parts of the proof that I was missing (IFS1 for E, and IFS4), and give them below. These proofs are probably difficult to follow without diagrams (at least they would be so for me!), so you may want to draw diagrams as you go through them. I'll try to add diagrams to the post once I figure out how to do it.

Both proofs rely on an earlier result stating that all coequalizers are epimorphisms.

IFS1 for E

For $i\in \{0, 1, 2\}$, let $e_i\in\mathbf{E}$ be a coequalizer for the fork $\alpha_i, \beta_i:\bullet \rightrightarrows \bullet$. Therefore, $e_i\;{\scriptstyle \circ}\;\alpha_i = e_i\;{\scriptstyle \circ}\;\beta_i$. Furthermore, suppose that cod($e_1$) = dom($e_2$), and that for some $m\in\mathbf{M}$, $\;e_2\;{\scriptstyle \circ}\;e_1 = m\;{\scriptstyle \circ}\;e_0$. Therefore we have $$ \begin{align} e_1\;{\scriptstyle \circ}\;\alpha_1 = e_1\;{\scriptstyle \circ}\;\beta_1 & \Rightarrow e_2\;{\scriptstyle \circ}\;e_1\;{\scriptstyle \circ}\;\alpha_1 = e_2\;{\scriptstyle \circ}\;e_1\;{\scriptstyle \circ}\;\beta_1\\ & \Rightarrow m\;{\scriptstyle \circ}\;e_0\;{\scriptstyle \circ}\;\alpha_1 = m\;{\scriptstyle \circ}\;e_0\;{\scriptstyle \circ}\;\beta_1\\ & \Rightarrow e_0\;{\scriptstyle \circ}\;\alpha_1 = e_0\;{\scriptstyle \circ}\;\beta_1 \end{align} $$ Therefore, by definition of $e_1$, there is a unique morphism $u$ such that $e_0 = u\;{\scriptstyle \circ}\;e_1$. Therefore, $e_2\;{\scriptstyle \circ}\;e_1 = m\;{\scriptstyle \circ}\;e_0 = m\;{\scriptstyle \circ}\;u\;{\scriptstyle \circ}\;e_1$. Since $e_1$ is epic, we have $m\;{\scriptstyle \circ}\;u = e_2$. Now, $$ \begin{align} e_2\;{\scriptstyle \circ}\;\alpha_2 = e_2\;{\scriptstyle \circ}\;\beta_2 & \Rightarrow m\;{\scriptstyle \circ}\;u\;{\scriptstyle \circ}\;\alpha_2 = m\;{\scriptstyle \circ}\;u\;{\scriptstyle \circ}\;\beta_2\\ & \Rightarrow u\;{\scriptstyle \circ}\;\alpha_2 = u\;{\scriptstyle \circ}\;\beta_2 \end{align} $$ Therefore, by definition of $e_2$, there is a unique morphism $v$ such that $u = v\;{\scriptstyle \circ}\;e_2$. It follows that $v\;{\scriptstyle \circ}\;m = 1_{\mathrm{cod}(e_0)}$ and $m\;{\scriptstyle \circ}\;v = 1_{\mathrm{cod}(e_2)}$, since $e_0$ and $e_2$ are coequalizers. Therefore, $m$ is an isomorphism. From this it follows that $\;e_2\;{\scriptstyle \circ}\;e_1 = m\;{\scriptstyle \circ}\;e_0$ is a coequalizer.

From the foregoing argument it follows that E is closed with respect to composition.


For $i\in \{0, 1\}$, let $m_i\in\mathbf{M}$, let $e_i\in\mathbf{E}$ be a coequalizer for the fork $\alpha_i, \beta_i:\bullet \rightrightarrows \bullet$, and suppose that $m_0\;{\scriptstyle \circ}\;e_0 = m_1\;{\scriptstyle \circ}\;e_1$.

Therefore we have $$\begin{align} e_0\;{\scriptstyle \circ}\;\alpha_0 = e_0\;{\scriptstyle \circ}\;\beta_0 & \Rightarrow m_0\;{\scriptstyle \circ}\;e_0\;{\scriptstyle \circ}\;\alpha_0 = m_0\;{\scriptstyle \circ}\;e_0\;{\scriptstyle \circ}\;\beta_0\\ & \Rightarrow m_1\;{\scriptstyle \circ}\;e_1\;{\scriptstyle \circ}\;\alpha_0 = m_1\;{\scriptstyle \circ}\;e_1\;{\scriptstyle \circ}\;\beta_0\\ & \Rightarrow e_1\;{\scriptstyle \circ}\;\alpha_0 = e_1\;{\scriptstyle \circ}\;\beta_0 \end{align} $$ Since $e_0$ is a coequalizer, there is a unique morphism $\psi$ such that $e_1 = \psi\;{\scriptstyle \circ}\;e_0$. (From this we also get that $m_0\;{\scriptstyle \circ}\;e_0 = m_1\;{\scriptstyle \circ}\;\psi\;{\scriptstyle \circ}\;e_0$, so $m_0 = m_1\;{\scriptstyle \circ}\;\psi$, since $e_0$ is epic.) By symmetry, we establish the existence of a unique morphism $\theta$ such that $e_0 = \theta\;{\scriptstyle \circ}\;e_1$ (and also $m_1 = m_0\;{\scriptstyle \circ}\;\theta$). Since both $e_0$ and $e_1$ are coequalizers, it follows that $\theta\;{\scriptstyle \circ}\;\psi = 1_{\mathrm{cod}(e_0)}$ and $\psi\;{\scriptstyle \circ}\;\theta = 1_{\mathrm{cod}(e_1)}$. Therefore, $\psi$ is the isomorphism that establishes IFS4.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.