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Let $A \xrightarrow{f} B$ be a morphism of objects $A,B$ of a category $\mathcal{C}$. Then

  • We say that a morphism $K \xrightarrow{\iota_K} A$ of objects of $\mathcal{C}$ is a $\mathit{kernel}$ of $f$ if $f\iota_K = 0$ and whenever $K' \xrightarrow{\iota_{K'}} A$ is another morphism satisfying $f\iota_{K'} = 0$ there exists a unique morphism $K' \xrightarrow{\mathscr{k}} K$ such that $\iota_{K'} = \iota_K\mathscr{k}$.
  • We say that a morphism $A \xrightarrow{e_I} I$ of objects of $\mathcal{C}$ is an $\mathit{image}$ of $f$ if there exists a monomorphism $I \xrightarrow{\varepsilon_I} B$ with $f = \varepsilon_I e_I$ such that whenever $A \xrightarrow{e_{I'}} I’$ is a morphism such there exists a monomorphism $I' \xrightarrow{\varepsilon_{I'}} B$ with $f = \varepsilon_{I'}e_{I'}$ there exists a unique morphism $I \xrightarrow{\mathscr{i}} I'$ such that $\varepsilon_I = \varepsilon_{I'}\mathscr{i}$.

My question concerns the directions of the arrows $\mathscr{k}, \mathscr{i}$. In the case of the kernel, $K$ is the codomain of $\mathscr{k}$ whilst in the case of the image, $I$ is the domain of $\mathscr{i}$. Is there a reason it is defined this way? And if so is there a good way to remember which is which? Note that the definition of cockerel/coimage reverses these too.

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  • $\begingroup$ Your map $e_I$ in the definition of the image should be a map $A\to I$, not $I\to A$. $\endgroup$ – Andrew Hubery May 9 at 21:28
  • $\begingroup$ @AndrewHubery too right! Fixing it $\endgroup$ – Adam Higgins May 9 at 21:29
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Kernel is an example of a limit, while image is an example of a colimit. Morphisms induced by universal properties always go into limit and out of a colimit.

You can easily tell why kernel has to be a limit: it's the largest subobject of domain of $f$ on which $f$ vanishes. So, if $f$ vanishes on something, there's an appropriate morphism going into kernel.

Similarly, image is colimit since it's the smallest subobject of codomain of $f$ through which $f$ factors. If there is any other such subobject, image is contained in it, i.e. there is an appropriate morphism from image.

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    $\begingroup$ Thank you! Absolutely perfect answer! Exactly what I was looking for $\endgroup$ – Adam Higgins May 9 at 21:05

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