User herman stel - MathOverflow

Answer by Herman Stel for Image of composite morphisms

2010-08-29T03:41:11Z

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 $\gamma(z)\vdash_{z:Z} \quad \exists y(\beta(y)\wedge g(y)=z)\vdash_{z:Z} \quad \exists x(\alpha(x)\wedge g(f(x))=z)$.

Answer by Herman Stel for Universal functors according to Cohn.

2010-08-15T22:54:07Z

What Cohn calls a representation is what Jacob Lurie calls a correspondence in Higher Topos Theory (which you can download from his website), in a subsection called "correspondences", which is 2.3.1 in my copy. Given a correspondence $C'^{op}\times C\rightarrow Set$ we can form the join along this correspondence, as you may read in Lurie's book. Then take "I" to be the inclusion into this join, and "U" to be as described in Cohn.

Answer by Herman Stel for Computing colimits in a Lawvere theory

2010-08-15T02:56:23Z

Let $T$ be a Lawvere theory. Then view the category of $T$-algebras, as Professor Blass wrote, as a class of universal algebras for a signature that corresponds to $T$ and is equationally definable, i.e. a variety. Then do what one does with these kinds of algebras:

To get coequalizers use quotients of algebras by what are called congruences (at least, that is what they are called in Universal Algebra). A congruence is an equivalence relation on the set that corresponds under your functor $A$ to the object 1 in $T$ (that is, the underlying set) that is preserved under all operations of the theory (arrows in $T$).

To get coproducts construct the left adjoint by defining the free $T$-algebra on some set $S$ to be the obvious structure on the set of all formal expressions $f(s_1,...,s_n)$ with $f$ an arrow of $T$ and $s_1,...,s_n\in S$. It is then clear what the coproduct of a set of free $T$-algebras is, and you can use congruences to get the coproduct on an arbitrary set of $T$-algebras.

Comment by Herman Stel

2010-08-16T07:28:59Z

Yes, this is the join in Lurie's book.