Quick version of my question: do all small colimits exist in the category of all groups? Of rings? Of modules? (If not, I expect to be able to take a counterexample and use it to answer by real question)(adsbygoogle = window.adsbygoogle || []).push({});

I'm reading through Jacobson's Basic Algebra II, and have a question about the section on colimits in a category of [itex]\Omega[/itex]-algebras.

The theorem Jacobson states is that if you have a diagram in [itex]\mathbf{\Omega}[/itex]-Algwhere for any two objects A and B in the diagram, the diagram also contains an object C and arrows [itex]A \rightarrow C[/itex] and [itex]B \rightarrow C[/itex], then the colimit exists.

He doesn't say anything about whether a more general colimit can fail to exist, so my question is if all small colimits exist, or if there are examples of diagrams where the colimit fails to exist.

I think I've proven that all colimits exist, but I'm kind of suspicious of my work and if true, curious why Jacobson didn't prove the general case...

Anyways, my proof is as follows:

Suppose we have a diagram indexed by the preorder I. We denote the objects by [itex]A_\alpha[/itex], and the arrows by [itex]\phi_{\alpha \beta}:A_\alpha \rightarrow A_\beta[/itex] for any [itex]\alpha \leq \beta[/itex].

The arrows satisfy [itex]\phi_{\beta \gamma} \circ \phi_{\alpha \beta} = \phi_{\alpha \gamma}[/itex] for any [itex]\alpha \leq \beta \leq \gamma[/itex].

Let X be the disjoint union of the elements of all of the [itex]A_\alpha[/itex], and let F be the free algebra [itex]F(\Omega, X)[/itex].

Let [itex]\Theta[/itex] be the congruence on F generated by the relations [itex]a \equiv \phi_{\alpha, \beta}(a)[/itex] for [itex]a \in A_\alpha[/itex], and by the relations given by the arithmetic of each [itex]A_\alpha[/itex].

I claim that [itex]L = F / \Theta[/itex] is a colimit of the diagram.

Let [itex]i_\alpha : A_\alpha \rightarrow L[/itex] be given by [itex]i_\alpha(a) = [a]_\Theta[/itex] (the equivalence class of a under the congruence [itex]\Theta[/itex]).

We have that [itex]i_\beta \circ \phi_{\alpha \beta} = i_\alpha[/itex] because:

[tex]i_\beta (\phi_{\alpha \beta} (a) ) = [ \phi_{\alpha \beta} (a) ]_\Theta[/tex]

[tex]i_\alpha (a) = [a]_\Theta[/tex]

and that [itex]a \Theta \phi_{\alpha \beta}(a)[/itex].

Now, let B be any [itex]\Omega[/itex]-algebra, and let there be maps [itex]\psi_\alpha:A_\alpha \rightarrow B[/itex] satisfying [itex]\psi_\beta \phi_{\alpha \beta} = \psi_\alpha[/itex].

Then, there exists a unique map [itex]f:F \rightarrow B[/itex] given by [itex]f(a) = \psi_\alpha(a)[/itex] for [itex]a \in A_\alpha[/itex].

All of the relations [itex]a \equiv \phi_{\alpha, \beta}(a)[/itex] for [itex]a \in A_\alpha[/itex] are in the kernel of f, and so are the relations given by the arithmetic of each [itex]A_\alpha[/itex], so f must factor uniquely through [itex]L = F / \Theta[/itex], giving a unique map [itex]p:F / \Theta \rightarrow B[/itex] such that [itex]p([a]_\Theta) = f(a) = \psi_\alpha(a)[/itex] for [itex]a \in A_\alpha[/itex].

We have [itex]p \circ i_\alpha = \psi_\alpha[/itex] because:

[tex]p( i_\alpha(a)) = p([a]_\Theta) = \psi_\alpha(a)[/tex]

Therefore, L is a colimit of the diagram.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Colimits vs algebra

Loading...

Similar Threads for Colimits algebra | Date |
---|---|

I Adding a matrix and a scalar. | Mar 31, 2018 |

I Linear mapping of a binary vector based on its decimal value | Mar 23, 2018 |

I Semi-simple Lie algebra | Mar 10, 2018 |

I Can we construct a Lie algebra from the squares of SU(1,1) | Feb 24, 2018 |

Colimits of topological spaces | Feb 19, 2007 |

**Physics Forums - The Fusion of Science and Community**