Timeline for In the category of sets epimorphisms are surjective - Constructive Proof?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 9, 2022 at 16:47 | answer | added | Toby Bartels | timeline score: 5 | |
| Aug 20, 2014 at 12:45 | vote | accept | Gerrit Begher | ||
| Aug 19, 2014 at 16:58 | history | edited | Max Horn | CC BY-SA 3.0 |
fix typo
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| Aug 18, 2014 at 19:50 | comment | added | Michal R. Przybylek | @AndreasCaicedo, I do not understand why you have removed tag "set-theory". As for me, the question is exactly about set theory (moreover, it is formulated in such a way, that it is about the standard set theory). | |
| Aug 18, 2014 at 18:57 | history | edited | Andrés E. Caicedo |
edited tags
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| Aug 18, 2014 at 18:32 | answer | added | Andrej Bauer | timeline score: 22 | |
| Aug 18, 2014 at 13:09 | answer | added | aws | timeline score: 8 | |
| Aug 18, 2014 at 12:04 | answer | added | Michal R. Przybylek | timeline score: 10 | |
| Aug 18, 2014 at 11:22 | comment | added | Simon Henry | I think that depend on what you mean by "constructive". if it is "topos valid"/without excluded middle then it is indeed true : consider two maps from $Y$ to the object obtained from $Y$ be identifying any two points which are in the image of $X$. | |
| Aug 18, 2014 at 10:44 | comment | added | Emil Jeřábek | @JochenWengenroth: Why is $y\in f(X)$ decidable? | |
| Aug 18, 2014 at 10:24 | comment | added | Jochen Wengenroth | Given the epimorphism $f$ define $g:Y\to\lbrace 0,1\rbrace$ by $g(y)=1$ if $y\in f(X)$ and $g(y)=0$ else. For the constant function $h(y)=1$ you have $g\circ f=h\circ f$ so that $g=h$. Hence, every $y\in Y$ belongs to $f(X)$ and $g$ is surjective. | |
| Aug 18, 2014 at 9:56 | history | asked | Gerrit Begher | CC BY-SA 3.0 |