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.

Suppose $F : C \to D$ is a functor between concrete categories. If $U_D : D \to \text{Set}$ and $U_C : C \to \text{Set}$ are the corresponding forgetful functors to $\text{Set}$, then it's natural to require that $U_C = U_D F$.

Suppose $U_C$ is representable. I would like to say then that $U_D$ is a lift of the representable functor $U_C$ to $D$, or something like that. The representing object should then acquire extra structure corresponding to whatever is needed for $F$ to land in $D$. Is there an established term for this?

share|improve this question
    
I'm not sure I understand the question. It is easy to see that the forgetful functor $U_{C}$ is represented by an object $c$ if and only if $c$ is free over the singleton set. Your requirement $U_{C} = U_{D}F$ says that $F$ is a concrete functor. It is also easy to see that concrete functors are faithful and completely determined by their object function (because $U_C$ and $U_D$ are faithful by definition). I cannot really interpret your intention to speak of a lifting property in this context. Could you elaborate a little? –  t.b. Jan 20 '11 at 15:56
    
@Theo: it's not really a lifting property. The representability is the part I care about. I want to say that the object that represents U_C also "represents" F in an appropriate sense, so I want to say that F is a "lifted" representable functor. Does that make sense? That's the concept I care about. (Also, in hindsight, I don't care whether F is forgetful or not.) –  Qiaochu Yuan Jan 20 '11 at 16:15
    
Isn't this similar to Yoneda's Lemma? –  PEV Jan 20 '11 at 17:21
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.