Timeline for Adjunctions (Reflection)
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23 at 21:24 | comment | added | Siya | Yes! that makes sense. That's what I was looking for. Many Thanks! | |
| Jul 23 at 21:22 | comment | added | Todd Trimble | Make sense? We have $\eta_Y g \eta_Y = \eta_Y$ using the fact that $g$ is a left inverse of $\eta_Y$. So $\eta_Y g \eta_Y = 1_Y \circ \eta_Y$. Now conclude $\eta_Y g = 1_Y$ since $\eta_Y$ is an epimorphism. | |
| Jul 23 at 21:14 | comment | added | Todd Trimble | Oh, I see what you're trying to ask (I was thrown off by your notation). Since you've shown $\eta_Y$ has a left inverse, it will suffice to show it is an epimorphism. So suppose $f, g$ are distinct morphisms of the form $Y \to Y'$. Since the diagonal arrow in your diagram is injective, the two morphisms $f \eta_Y, g\eta_Y$ of the form $LRY \to Y'$ are also distinct. And this completes the proof. | |
| Jul 23 at 20:57 | comment | added | Siya | Alright. I guess the approach I used with showing $g$ is also a right inverse is not needed? | |
| Jul 23 at 20:31 | history | edited | LSpice | CC BY-SA 4.0 |
Re-ordering so that picture doesn't occur mid-sentence
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| Jul 23 at 20:23 | comment | added | Todd Trimble | When you say $R$ is fully faithful, that means the arrow you've denoted by $R$ is injective (that's the faithful part) and surjective (that's the full part). So this arrow $R$ is bijective: is invertible. So $\eta_Y$ is a composition of two isomorphisms. | |
| Jul 23 at 20:18 | history | asked | Siya | CC BY-SA 4.0 |