# Map factorization in an abelian category

• Jan 5th 2012, 02:37 AM
albi
Map factorization in an abelian category
Suppose we are working in an abelian category. Prove that any map $f: A \rightarrow B$ has a decomposition $A \rightarrow \mathrm{im}(f) \rightarrow B$ where the left arrow is epi and the right arrow is monic.

My attempt of solution

Let's consider cokernel of f: $\mathrm{coker}(f): B \rightarrow C$ it exists since we are in abelian category. For the same reason the cokernel has a kernel $\mathrm{ker coker}(f): \mathrm{im}(f) \rightarrow B$.

Since $\mathrm{coker}(f) \circ f = 0$ there is map $f': A \rightarrow \mathrm{im}(f)$ such that $f = \mathrm{ker coker}(f) \circ f'$. This is the factorization we were looking for.

Immediatly we see, that the right arrow is monic (since kernels are monics), so the only problem is to prove that $f'$ is epi. And this is the thing i cannot do :(

I was unable to prove that $\mathrm{coker}(f) = 0$ since i cannot see any nonzero arrows from $\mathrm{im}(f)$ (and I need such arrow in order to use cokernel universality property).

Yet I was able to see some things...

Let $m = \mathrm{ker coker}(f)$. One thing I was able to notice is that $\mathrm{coker}(m) = coker(f)$. It is obvious that we have to use property, that the rightmost arrow is a cokernel (we can do the above construction with any map for which composition with f is 0, but f' need not to be epi). However again I don't see any arrows from B (so i cannot use cokernel property).

I had some other idea to deal with it. Let us start "from the other side". So consider kernel of f: $\mathrm{ker}(f): K \rightarrow A$. The analogous construction yields decomposition $A \rightarrow \mathrm{cim}(f) \rightarrow B$ where the first arrow is $\mathrm{coker ker}(f)$ and the second one is denoted by $f''$.

I was able to construct a map $i: \mathrm{cim}(f) \rightarrow \mathrm{im}(f)$ which has good "composition properties" (the corresponding triangles are commutative), however i was unable to prove that $i$ is an isomorphism (i was unable to construct the arrow in the opposite direction).

I would appreciate any ideas concerning this problem.
• Jan 17th 2012, 08:01 AM
albi
Re: Map factorization in an abelian category
It looks like it is the uniqueness part I was missing all the time.

I have found the proof in here 3xw.tac.mta.ca/tac/reprints/articles/3/tr3.pdf (3xw = www)
It is lucid, however not direct (not in a way I think about abelian category) so i will present my version of this proof (which is more complicated)
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Step 0

Let us start with $f: A \rightarrow B$, let $c: B \rightarrow C$ denote cokernel of f.

We know that f factors through $f': A \rightarrow \mathrm{im}(f)$ and $\mathrm{ker}(c): \mathrm{im}(f) \rightarrow B$.

We can do the same with f'. Namely let $c': \mathrm{im}(f) \rightarrow C'$ denote cokernel of f'. Then f' factors through $f'': A \rightarrow \mathrm{im}(f')$ and $\mathrm{ker}(c'): \mathrm{im}(f') \rightarrow \mathrm{im}(f)$.

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Step 1
There is a monomorphism $k: \mathrm{im}(f) \rightarrow \mathrm{im}(f')$

Proof:
Let $m = \mathrm{ker}(c) \circ \mathrm{ker}(c')$. Then we can see that there is a map e such that $\mathrm{coker}(m) = e \circ c$.

We can see that $\mathrm{coker}(m) \circ \mathrm{ker}(c) = e \circ c \circ \mathrm{ker}(c) = 0$. Hence $\mathrm{ker}(c)$ factorize through m. So we have $\mathrm{ker}(c) = \mathrm{ker}(c') \circ k \circ \mathrm{ker}(c)$.

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Step 2

Since $\mathrm{ker}(c)$ is monomorhpism we have $\mathrm{ker}(c') \circ k = \mathrm{id}$.

Next we see that $\mathrm{ker}(c') \circ k \circ \mathrm{ker}(c') = \mathrm{ker}(c')$ so we can see that $k \circ \mathrm{ker}(c') = \mathrm{id}$. Therefore $\mathrm{ker}(c')$ is an isomorphism.

Finally $0 = \mathrm{coker}( \mathrm{ker}(c')) = c'$ because every epi is cokernel of its kernel in abelian category.

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My remarks.

I wouldn't be able to do this proof without the book. Their aproach is more lucid, however it is not so beautyful in my opinion.

They argument that there should be not proper subobject $\mathrm{im}(f')$ in $\mathrm{im}(f)$ because image can be thought as the smallest subobject that factors f.

It is a good idea to think about image that way. Conversely cokernel can be thought as the biggest quotient object.

Also it is a good idea to think about Ker and Coker as inverse maps on quotient objects and subobjects. And what is important they are order reversing (step one above is the construction of that).