I am new to categories and I found in a book that it is possible to construct a category in which the following are true: there exist morphisms $f:A \to B$ and $g:B \to C$, and monomorphisms $\alpha:A' \to A$, $I:B' \to B$ and $J:C' \to C$ such that

(1) $I$ is an image of $\alpha$ under $f$.

(2) $J$ is an image of $I$ under $g$.

(3) $J$ is not an image of $\alpha$ under $g \circ f$.

This is rather counterintuitive, but I kind of see that what is counterintuitive is the definition of "image".

So, here comes my question: What are conditions to impose on the category to ensure that (3) does not occur?


This cannot happen in a regular category. Below I give a proof using the sequent calculus of subobjects in a regular category. It can be deciphered using the book 'Sketches of an Elephant Volume 2' by Peter T. Johnstone, in particular chapter D1.

I write $\beta:=I$ and $\gamma:=J$. I hope the definition of image given in your book is the same as mine, namely the image of a subobject (~mono) $S$ under a morphism $\phi$ is the least subobject of the codomain of $\phi$ through which $\phi\circ\overline{S}$ factors, where $\overline{S}\in S$.

Assume we know $\exists x(\alpha(x)\wedge f(x)=y) \dashv\vdash_{y:Y}\quad \beta(y)$ and $\exists y(\beta(y)\wedge g(y)=z) \dashv\vdash_{z:Z}\quad \gamma(z)$. We then want to prove two things. The first is that $\exists x(\alpha(x)\wedge g(f(x))=z)\vdash_{z:Z}\quad \gamma(z)$, the second that $\gamma(z)\vdash_{z:Z} \quad \exists x(\alpha(x)\wedge g(f(x))=z)$.

For the first we have the following. $\alpha(x)\wedge g(f(x))=z$ $\vdash_{x:X,z:Z} \quad\alpha(x)\wedge g(f(x))=z \wedge f(x)=f(x)$ $\vdash_{x:X,z:Z}\quad \alpha(x)\wedge g(f(x))=z \wedge \beta(f(x))$ $\vdash_{x:X,z:Z}\quad \gamma(g(f(x)))$. Therefore $\alpha(x)\wedge g(f(x))=z\vdash_{x:X,z:Z}\quad \gamma(z)$ and hence $\exists x(\alpha(x)\wedge g(f(x))=z)\vdash_{z:Z}\quad \gamma(z)$.

The second also holds. First note that $\beta \wedge g(y)=z$ $\vdash_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge f(x)=y)\wedge g(y)=z$ $\vdash_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge f(x)=y\wedge g(y)=z)$ $\vdash_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge g(f(x))=z)$ from which we may conclude that $\gamma(z)\vdash_{z:Z} \quad \exists y(\beta(y)\wedge g(y)=z)\vdash_{z:Z} \quad \exists x(\alpha(x)\wedge g(f(x))=z)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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