# The Unapologetic Mathematician

## Images, Coimages, and Exactness

By the first isomorphism theorem, we know that any morphism $f$ in an abelian category $\mathcal{C}$ factorizes as $f=m\circ e$ with $m=\mathrm{Ker}(\mathrm{Cok}(f))$, and $e$ is epic. Since $m$ is monic, $f\circ t=m\circ e\circ t=0$ exactly when $e\circ t=0$. That is, the kernel of $f$ is isomorphic to the kernel of $e$. Then since $e$ is epic, $e=\mathrm{Cok}(\mathrm{Ker}(e))=\mathrm{Cok}(\mathrm{Ker}(f))$. So there’s a sort of a symmetry here between the monic and the epic in the factorization of $f$.

Now let’s consider another morphism $f'$ and a pair of morphisms $(g,h)$ so that $h\circ f=f'\circ g$. Then we can factorize each of $f$ and $f'$ as above to find $h\circ m\circ e=m'\circ e'\circ g$. Then there is a unique $k$ such that $e'\circ g=k\circ e$ and $m'\circ k=h'\circ m$.

To see this, set $u=\mathrm{Ker}(f)=\mathrm{Ker}(e)$. Then $0=h\circ f\circ u=m'\circ e'\circ g\circ u$ so $e'\circ g\circ u=0$. Thus $e'\circ g$ factors uniquely through $e=\mathrm{Cok}(u)$ as $e'\circ g=k\circ e$. Then $m'\circ k\circ e=m'\circ e'\circ g=h\circ m\circ e$. And so since $e$ is epic we have $m'\circ k=h'\circ m$.

Now, we’ll regard $f$ and $f'$ as objects in the arrow category $\mathcal{C}^\mathbf{2}$. Then the pair $(g,h)$ is a morphism from $f$ to $f'$. Similarly, the triangle $f=m\circ e$ is an object of $\mathcal{C}^\mathbf{3}$, and the triple $(g,k,h)$ is a morphism in this category.

What the above proof shows is that any object in $\mathcal{C}^\mathrm{2}$ can be assigned an object in $\mathcal{C}^\mathcal{3}$, and that any morphism in $\mathcal{C}^\mathbf{2}$ can be assigned one in $\mathcal{C}^\mathcal{3}$. Clearly this assignment amounts to a functor. In particular, if we start with the identity pair $(1,1)$ we must have an isomorphism for $k$, and thus any two factorizations are isomorphic.

Now, given this unique (up to isomorphism) factorization, we can define the image and coimage of $f=m\circ e:A\rightarrow B$ as $\mathrm{Im}(f)=m$ and $\mathrm{Coim}(f)=e$. Thus as expected the image of $f$ is a subobject of its target, and the coimage is a quotient object of its source.

Now that we have defined images and coimages we can define what it means for a composable sequence of morphisms to be exact. Let’s say we have $f:A\rightarrow B$ and $g:B\rightarrow C$. Both $\mathrm{Im}(f)$ and $\mathrm{Ker}(g)$ are subobjects of $B$, and we say that the pair $(f,g)$ is exact at $B$ when $\mathrm{Im}(f)=\mathrm{Ker}(g)$. We say that a longer string of composable arrows is exact if it is exact at each object inside the string.

As a special case, we say the sequence $\mathbf{0}\rightarrow A\rightarrow B\rightarrow C\rightarrow\mathbf{0}$ is short exact if it is exact. That is, if we let the two outer arrows be the unique such, let $f:A\rightarrow B$, and let $g:B\rightarrow C$, then the sequence is short exact if $\mathrm{Ker}(f)=\mathbf{0}$, $\mathrm{Im}(g)=\mathbf{0}$, and $\mathrm{Ker}(g)=\mathrm{Im}(f)$. If we drop the left $\mathbf{0}$ we call the sequence short right exact, and short left exact sequences are defined similarly.

Now the factorization of $f:A\rightarrow B$ gives rise to two short exact sequences: $\mathbf{0}\rightarrow\mathrm{Ker}(f)\rightarrow A\rightarrow\mathrm{Coim}(f)\rightarrow\mathbf{0}$ and $\mathbf{0}\rightarrow\mathrm{Im}(f)\rightarrow B\rightarrow\mathrm{Cok}(f)\rightarrow\mathbf{0}$. Then because the objects of the coimage and the image are isomorphic, we can weave these two sequences together at that point. In fact, we did something just like this back when we talked about exact sequences of groups!

An $\mathbf{Ab}$-functor $T:\mathcal{C}\rightarrow\mathcal{D}$ is called left exact when it preserves all finite limits. In particular it preserves kernels — that is, left exact sequences. Since any $\mathbf{Ab}$-functor preserves biproducts, preserving kernels is enough to preserve all finite limits. Similarly, a right exact functor is one which preserves all finite colimits, or equivalently all cokernels — right exact sequences. Finally, a functor is exact if it is both left and right exact.

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Posted by John Armstrong | Category theory

## 4 Comments	»

1. [...] Exact Sequences Last time we defined a short exact sequence in an abelian category to be an exact sequence of the form . [...]

Pingback by | September 27, 2007 | Reply

2. [...] So let’s take this and consider a linear transformation . The first isomorphism theorem says we can factor as a surjection followed by an injection . We’ll just regard the latter as the inclusion of the image of as a subspace of . As for the surjection, it must be the linear map , just as in any abelian category. Then we can set up the short exact sequence [...]

Pingback by | June 27, 2008 | Reply

3. [...] all the machinery of homological algebra, if we should so choose. In particular, we can talk about exact sequences, which can be useful from time to time. Possibly related posts: (automatically [...]

Pingback by | December 15, 2008 | Reply

4. [...] are the morphisms in the category of -modules. It turns out that this category has kernels and has images. Those two references are pretty technical, so we’ll talk in more down-to-earth [...]

Pingback by | September 29, 2010 | Reply

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