Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

An initial object of a category $\mathcal{C}$ is an object $I$ in $\mathcal{C}$ such that for every object $X$ in $\mathcal{C}$, there exists precisely one arrow $I → X$.

Let $\mathcal{X}$ and $\mathcal{A}$ be categories and let $U:\mathcal{X} \rightarrow \mathcal{A}$ be a functor. Let $A$ be an object of $\mathcal{A}$ and define $(A\downarrow U)$ to be the category with

  • Objects: pairs $\langle X, h: A \rightarrow UX \rangle$ where $X$ is an object of $\mathcal{X}$ and $h$ is an arrow of $\mathcal{A}$
  • Arrows: $f: \langle X, h: A \rightarrow UX \rangle \rightarrow \langle X', h': A \rightarrow UX' \rangle$ given by arrows $f: X \rightarrow X'$ of $\mathcal{X}$ such that $(Uf)\circ h = h'$ in $\mathcal{A}$.

Question: If there is no initial object in the category $\mathcal{X}$ can there still be an initial object in the category defined above?

share|improve this question
1  
Yes, for the case that $\mathcal{A}$ is the category of one object and exactly one morphism. I'm sure there are also more interesting examples out there. –  Shaun Ault Apr 10 '12 at 12:15
add comment

2 Answers

up vote 0 down vote accepted

Certainly $A\downarrow U$ can have an initial object without $\cal X$ having one. For example, take ${\cal A} = {\cal X}$ and let $U:{\cal X}\to{\cal A}$ be the identity functor. Then $(A,id_A)$ is the initial object.

On the other hand, even if $\cal X$ has an initial object, that cannot guarantee the existence of an initial object in $A\downarrow U$, for every object $A$, because this would imply that $U$ has a left adjoint.

share|improve this answer
add comment

I don't understand Shaun's example. Nevertheless, it is easy to think of other counterexamples. Take $\mathcal X = \mathcal A$ the category with two objects, two identity morphisms and no other maps, and let $U$ be the identity functor. Then $\mathcal X$ has no initial object, but the comma category is the one-object one-morphism category, which does have one.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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