Timeline for Pullback and pseudoelements
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 8, 2022 at 16:20 | comment | added | Ricky | Thanks for the answer, I am checking the details. Just a typo: in the two diagrams with $\eta$, the bottom row is $f^\ast$, and in the following diagram the curved arrow goes to $B$. | |
| Apr 8, 2022 at 16:08 | history | edited | user473423 | CC BY-SA 4.0 |
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| Apr 8, 2022 at 14:27 | history | edited | user473423 | CC BY-SA 4.0 |
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| Apr 8, 2022 at 14:02 | comment | added | user473423 | Sorry, you are right. But indeed, I do not use this point in the proof. | |
| Apr 8, 2022 at 13:58 | history | edited | user473423 | CC BY-SA 4.0 |
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| Apr 8, 2022 at 13:06 | comment | added | Ricky | This indeed looks false: take $p : X \to P$, and define $p'$ by precomposing with an automorphism (different from the identity) of $X$. Then $p \neq p'$ but $[p] = [p']$. | |
| Apr 8, 2022 at 12:09 | comment | added | Ricky | Why you can assume the epi is the same? The definition of the equivalence relation is that there is an epi for $p$ and one for $p'$, having always the same source. Here they also have the same target, but they can be different. | |
| Apr 8, 2022 at 12:07 | comment | added | user473423 | If p and p' have the same domain, then [p]=[p'] means that there is an epi F such that $p\circ F=p'\circ F$. By the definition of an epimorphism this measn that $p=p'$. | |
| Apr 8, 2022 at 12:05 | comment | added | user473423 | I disagree, $f^*([p])=f^*([p'])$ means by definition that $f\circ p$ and $f\circ p'$ define the same element, that is, there exists an epi $d$ with $f\circ p\circ d=f\circ p'\circ d$ and that's exactly what the diagram says. | |
| Apr 8, 2022 at 11:49 | comment | added | Ricky | Also, I don't see why the diagram with $D$ and $D'$ is commutative. The only thing you can deduce from $f^\ast([p]) = f^\ast([p']) $ is that $f^\ast \circ p = f^\ast \circ p'$. You can play a similar game with $g^\ast$, but my point it exactly that it seems we have enough equations to finish the proof because usually all the diagram we can draw are commutative, but in this case it seems trickier. | |
| Apr 8, 2022 at 11:37 | comment | added | Ricky | I am sorry to be pedantic, but I don't understand all the details. First of all, why $[a]=[b]$ implies $a=b$ if $a$ and $b$ have the same codomain? | |
| Apr 8, 2022 at 7:30 | history | answered | user473423 | CC BY-SA 4.0 |