🪴 Quartz 3

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(co)limits do not respect homotopy equivalence

Last updated January 27, 2022

For example, we can regard $S^n$ as the colimit $$D^n\sqcup_{S^{n-1}}D^n\cong S^n.$$ Now $D^n\simeq \ast$, but $$\ast\sqcup \ast\not\simeq S^n.$$


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