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 you have a morphism $f:\mathcal{C}\to \mathcal{D}\times\mathcal{E}$ of categories. Then $f=\langle \pi_{\mathcal{D}}\circ f,\pi_{\mathcal{E}}\circ f\rangle$, uniquely, by the definition of the operation $\langle -,-\rangle$.

If $c\in\text{ob}\;\mathcal{C}$, can we say something about $f(c)$, in terms of this "decomposition"? (Suppose that we know how the objects behave under $\pi_{\mathcal{D}}\circ f$ and $\pi_{\mathcal{E}}\circ f$.)

share|improve this question
    
What do you mean "can we say something"? It will be a pair $(\pi_D \circ f(c), \pi_E \circ f(c))$ where $\pi_D \circ f(c) \in \operatorname{ob} \mathcal{D}$ and $\pi_E \circ f(c) \in \operatorname{ob} \mathcal{E}$. On objects a functor is just a map in the traditional sense, so if you understand maps from a set to a product of two sets, you're golden. –  Najib Idrissi Apr 22 at 14:09
    
Say $\pi_{\mathcal{D}}\circ f=F$ and $\pi_{\mathcal{E}}\circ f=G$. What I mean is, if we know what $F(c)$ and $G(c)$ are, can we say something about $f(c)$? –  Niels.Remb05 Apr 22 at 14:16
2  
Yes, you can say $f(c) = (F(c), G(c))$. I think you're a bit confused about the definition of the product category. The objects of $\mathcal{D} \times \mathcal{E}$ are pairs of objects of $\mathcal{D}$ and $\mathcal{E}$. –  Najib Idrissi Apr 22 at 14:17
    
Arghh I have been trying to do Exercise 3 of Chapter IV.1 in MacLane's CWM. I got very confused by trying to make sense of all the capital $C$ in the diagram. Shouldn't they be small $c$ everywhere?!...(Oh, I just noticed that maybe you don't have a copy of CWM?) –  Niels.Remb05 Apr 22 at 15:46
1  
Hmm...But $\delta_c$ is defined to be the unit of the adjunction $\Delta \dashv \times$. So shouldn't the letters in the diagram be small (not capitals)? –  Niels.Remb05 Apr 23 at 8:09

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.