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Jul 8, 2019 at 1:14 answer added Adrien Vakili timeline score: 1
Apr 13, 2017 at 12:58 history edited CommunityBot
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Feb 28, 2017 at 22:27 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jan 29, 2017 at 21:43 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Dec 30, 2016 at 20:55 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Nov 30, 2016 at 16:10 answer added David White timeline score: 2
Nov 30, 2016 at 16:10 comment added David White Note that using the "@" here will not send a notification to Neil Strickland, since he has nothing to do with this post.
Nov 30, 2016 at 14:01 comment added kousaka @user337830 Your explanation is very understandable, thanks a lot!!!
Nov 30, 2016 at 12:47 comment added user337830 The universal property used here is that of the functor $L_S\times L_S:\mathcal{M}\times\mathcal{M}\to \mathcal{M}[S^{-1}]\times\mathcal{M}[S^{-1}]$.
Nov 30, 2016 at 12:45 comment added user337830 You may would like to ask this question on math.stackexchange instead. Anyway, the assumption that $S$ is closed with respect to the monoidal product $\otimes$ implies that the composition of the localisation functor $L_S:\mathcal{M}\to \mathcal{M}[S^{-1}]$ with $\otimes$ sends a pair morphisms $(s,t)$ in $S\times S$ to an isomorphisms in $\mathcal{M}[S^{-1}]$. Thus, the composition $L_S\circ \otimes$ factorises through $L_S\times L_S$, defining the desired monoidal product on $\mathcal{M}[S^{-1}]\times\mathcal{M}[S^{-1}]$.
Nov 30, 2016 at 10:23 comment added kousaka I would appreciate it a lot if you could favour to answer this question..@Neil Strickland
Nov 30, 2016 at 10:22 review First posts
Nov 30, 2016 at 10:58
Nov 30, 2016 at 10:18 history asked kousaka CC BY-SA 3.0