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.

We want to characterize the limit and colimit of a functor $D\colon J\to \mathcal C$ by generalized elements. The existence of limits theorem states that the limit of $D$ is the equalizer of $$s,t\colon \prod_{j\in Ob(J)}D(j)\to \prod_{f\in Ar(J)}D(\text{cod}(f))$$ where $s=\langle Df\circ\pi_{\text{dom}(f)} \rangle_{f\in Ar(J)}$ and $t=\langle \pi_{\text{cod}(f)}\rangle_{f\in Ar(J)}$. Equalizers are monic, and so we may think of the limit of $D$ as a subobject of the product $\prod D(j)$.

Moreover, the Yoneda embedding $y\colon \mathcal C\to Set^{\mathcal C^{op}}$ preserves all limits. Therefore, it is tempting to write the limit of $D$ by generalized elements (i.e. $x\in X$ means $x\in Ar(\mathcal C)$ with cod$(x)$=X) as $$ \lim D = \left[x\in\prod D(j)| s(x)=t(x)\right]. $$

Unfortunately, the Yoneda embedding does not preserve colimits. However, the colimit of $D$ is a quotient object of the coproduct and I still feel tempted to write, $$ \text{colim } D = \frac{\coprod D(j)}{\sim}, $$ where $\sim = \bigcap [E:\text{equivalence relation} |\ \langle s(x), t(x)\rangle\in E]$ where the equivalence relations are taken on hom-sets. (And $s, t$ here are the dualized arrows of the maps above).

Is this the wrong characterization? Is there a good analogy between the colimit of an arbitrary functor and the colimit in sets? Is $$ \frac{\mathcal C(\square,\coprod D(j))}{\sim}\colon \mathcal C^{op}\to Set $$ a well-defined functor which is representable, i.e., isomorphic to $\mathcal C(\square,\text{colim } D)$?

share|improve this question
1  
I see this is the typical meaning of $\langle f \rangle_{f \in Ar(J)}$, and the notation for what I was looking for is $\prod_{f \in Ar(J)} f$ or $f \times g : X \times Y \to Z \times W$ for $f: X \to Z, g: Y \to W$. –  Loki Clock Mar 14 at 6:05
1  
Exactly. Moreover, the indexing sets have different sizes so such a product of arrows doesn't work here. –  Rachmaninoff Mar 14 at 6:07
    
The indices of the product range over $j\in Ob(J)$. –  Rachmaninoff Mar 14 at 6:28
    
Okay, so it's like $\pi_1:X \times Y \to X,$ not $\pi_X:X \times Y \to X$. –  Loki Clock Mar 14 at 6:30
    
Why the intersection over equivalence relations instead of the singleton/complement relation? –  Loki Clock Mar 14 at 6:45

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.