The proof of the following fact is trivial to most Arrow theorists / Linear algebraists, but I'm developing software that needs to "understand" in a sense this basic fact, because it is called upon all the time, so should be provable to the users of the software, in the software itself.

We are given the following diagram and the user is told that it commutes by definition.

g of f commuting triangle equaling the zero morphism

The goal diagram is the following:

Ker g contains image space of f

Using only the rules:

  1. Definition of kernel / image: What should we use here?
  2. Gluing two CD's along a common subdiagram and the resulting diagram is a CD.
  3. Deleting arrows in a CD and the diagram will still be a CD.
  4. You can compose along two paths in a CD with common endpoints and the resulting arrows are equal.
  5. Any others we require.

So I'm wondering about the definition of Kernel / Image. Clearly I'm using the one in which kernel / image are spaces, not maps. What is the most light-weight, yet most diagrammatic definitions of these.

Finally, how would the diagrammatic proof go given that, for example, the diagrams are of $R$-modules and $R$-module map?.

To create a visual diagram I used https://q.uiver.app. From there, create the diagram, turn off the grid and take a screenshot of it with Windows + Shift + S (or similar on other OS's), selecting a small region containing just the diagram, and then paste into the insert-image tool here.

Note: in the system language, dashed arrow means "there exists" whereas a normal arrow means "for all such" if it doesn't appear in surrounding text or it is simply the exact thing that the surrounding text talks about. So in the first diagram, $f,g$ both appear in the text, so they are "bound" whereas other variables might be "free to roam".

Edit. I got the vocabulary of "free" vs "bound" variable backwards. A bound variable is either a "there exists" or a "for all" variable for example (there are other variable-binding operators), where as a free variable is the one that occurs in the text for example "Definition of equalizer morphism $e$ is ..." so in the ... $e$ would be free. We can think of this nicely as bound variables are introduced inside the "box of a diagram" whereas free variables are "free from the box" and are typically introduced outside of a diagram.

  • 1
    $\begingroup$ How far do you want to go? Abelian categories, exact categories or even extriangulated? The definition in abelian categories is probably what you are looking for here $\endgroup$ Sep 18 at 21:10
  • 1
    $\begingroup$ Then if you use the definition of kernel as a limit it is direct but it may be considered as cheating? $\endgroup$ Sep 18 at 21:12
  • 1
    $\begingroup$ @julio_es_sui_glace you could do that, but that's bringing in a lot of machinery. I'm looking for more of a beginner-friendly analysis. After all we're just trying to prove a definition! So assume that I haven't yet implemented what a limit is or how diagrams behave under limits, etc. $\endgroup$ Sep 18 at 21:14
  • 2
    $\begingroup$ For me, a somewhat "canonical" argument would look something like: $\operatorname{im}$ is defined in terms of an epi-mono factorization system, so you have a factorization $A \twoheadrightarrow \operatorname{im}(f) \hookrightarrow B$. Then, composing with $g$ gives zero; so by the definition of kernel, you have a factorization $A \to \ker(g)$. And then, by epimorphisms being left-orthogonal to monomorphisms, you have a factorization $\operatorname{im}(f) \to \ker(g)$. That morphism is a mono since its composition with $\ker(g) \to B$ is mono. (Yes, probably more complex than you wanted.) $\endgroup$ Sep 18 at 21:18
  • 2
    $\begingroup$ Or, you could also use that if $A \twoheadrightarrow B \to C$ composes to 0, then the $B \to C$ map is 0; so in our situation $\operatorname{im}(f) \hookrightarrow B \to C$ is 0, giving a map $\operatorname{im}(f) \to \ker(g)$, and as before, use a composition being mono to convert this to a mono. $\endgroup$ Sep 18 at 21:44

2 Answers 2


$$g(f(x))=0 \rightarrow f(x)\in\ker(g) \rightarrow \left\{ f(x)|x-is-in-domain\right\}\subset \ker(g)$$ $$Im(f)\subseteq \ker(g)$$

  • $\begingroup$ You should use $x \in \text{dom } f$. But really this is not diagrammatic at all :| $\endgroup$ Sep 19 at 18:36
  • $\begingroup$ @MathCrackExchange how do you write in the text font and not in math font? $\endgroup$ Sep 19 at 20:42
  • 1
    $\begingroup$ I will teach you. Right click on my math in my comment (or anywhere on MSE) and goto "Show Math As..." > "TeX Commands". $\endgroup$ Sep 19 at 20:45

Thanks to Daniel Schepler in the comment section above, we have:

Epi-mono factorization of a morphism

as the definition of $\operatorname {im}f$. I'm assuming there is only one such, which makes sense because otherwise $f'$ would not essentially be equal to $f$ and so on..

Now considering the usual definition of kernel $(\ker g = K) \xrightarrow{k} Y$ and applying that UMP to the following situation works:

Complicated diagram involving kernel of g as an equalizer with zero and an epi-mono factorization of f

Though for the sake of the user I would definitely break this up into more steps.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .