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Feb 13, 2019 at 4:59 vote accept Praphulla Koushik
Feb 13, 2019 at 3:32 comment added Praphulla Koushik @MikeMiller Ofcourse they are surjective... Given $a\in \mathcal{G}_0$ I can just take $1_a:a\rightarrow a$ in $\mathcal{G}_0$ that says both $s,t$ are surjective... It was asked 5 hours ago i.e., 4 am for me.. I should have just slept off.. My mind did not work and I did not realise that then... I now feel like a real stupid :D I was also rechecking the definition in arxiv.org/pdf/math/0203100.pdf and it says it is just submersion :D :D It is already surjective so they just did not mention :P :P
Feb 13, 2019 at 1:30 history edited David Roberts CC BY-SA 4.0
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Feb 13, 2019 at 1:29 answer added David Roberts timeline score: 2
Feb 12, 2019 at 23:05 comment added mme If $s$ and $t$ are not surjective then $\mathcal G$ is not a category (and in particular not a groupoid), because every object in a category has an identity morphism.
Feb 12, 2019 at 21:55 history edited Praphulla Koushik CC BY-SA 4.0
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Feb 12, 2019 at 21:45 history asked Praphulla Koushik CC BY-SA 4.0