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Given a category $C$ with limits and colimits. Let $F: D \to C$ be a connected diagram in $C$. I want to show that if we just take the iterated pushouts then I will eventually get the colimit. For example if my diagram was $x \leftarrow u \rightarrow y \leftarrow v \rightarrow z$, then by taking the pushout $x \cup_u y$ and $y \cup_v z$. And then take the pushout $(x \cup_u y) \cup_y (y \cup_v z)$ this gives me the colimit of the diagram. With a small diagram like this it is easy to show but how would I go about showing this is true in general. I know by the Yoneda embedding and the Kan extension, I can just deal with presheafs. We can view each object in the diagram as a presheaf by the Yoneda embedding $Y : D \to \text{Set}^{D^{op}}$.

So now I just need to deal with a diagram of presheafs. The colimit of a diagram is just a weighted colimit when the weight is 1 on every object in the diagram. I want to show that the colimit of these presheafs is exactly the the presheaf that sends every object to 1. Sorry if I'm not clear anywhere. I'm quite new to Category theory.

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What you claim cannot possibly be true: how do you use just pushouts to construct the colimit of a diagram with more than one connected component? – Zhen Lin Nov 25 '13 at 9:48
Also you should write $x \sqcup_u y$ or $x \cup_u y$, not $\otimes$. – Martin Brandenburg Nov 25 '13 at 10:21
I forgot to add that it's connected. I want this to be true. So in a nice diagram I think it should be true, whatever "nice" means. – toopham Nov 25 '13 at 10:23
You should also allow infinite iterations. – Martin Brandenburg Nov 25 '13 at 11:01
Or otherwise restrict to finite connected diagrams. Then it is true: the colimit of a finite connected diagram can indeed be constructed by iterated pushouts. The proof is ugly, however. – Zhen Lin Nov 25 '13 at 11:18

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