7
$\begingroup$

I am trying to understand why we have a coequalizer $\sqcup_{0 \leq i < j \leq n} |\Delta^{n-2}| \rightrightarrows \sqcup_{0 \leq i \leq n} |\Delta^{n-1}| \rightarrow |\partial \Delta^n|$. What are all the three maps?

$\endgroup$
8
$\begingroup$

I realize I might be late but here are some hints. Let us look at simplicial sets instead of the realization, since the realization functor preserves colimits (and thus coequalizers) it will be enough. We have maps $d^i : \Delta^{n-2} \rightarrow \Delta^{n-1}$ as usual, and now, for each $i < j$ you have maps $d^jd^i = d^id^{j-1} : \Delta^{n-2} \rightarrow \Delta^n$ . The coequalizer is generated by these maps (induced by various coproducts and such) and the fact that $\partial \Delta^n$ is generated by $d^i:\Delta^{n-1} \rightarrow \Delta^n.$ ( the i-th faces) To see that this really is a coequalizer, you could aruge along the following lines. This is sketchy, but I think the idea is here. If we have a map $f: \amalg_{i=0}^n \Delta^{n-1} \rightarrow X$ such that for all $i<j$ $fd^i = fd^{j-1}$ you have to show that there's an unique map $h: \partial \Delta^n$ such that $hd^jd^i = hd^i d^{j-1} = fd^i = fd^{j-1}$. It should be easy to see what h must be. Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.