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May
8 |
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awarded | Caucus |
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Mar
27 |
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answered | Right-adjoint functors are left-exact? |
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Mar
15 |
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comment |
Image of Homomorphism of Lie groups Thanks. For your last paragraph, I can show that Lie group homomorphism has constant rank, and that injective map with constant rank must be an immersion (these facts are propositions in Lee's book). |
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Mar
15 |
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accepted | Image of Homomorphism of Lie groups |
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Mar
14 |
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comment |
Image of Homomorphism of Lie groups If you would please post your comment as an answer, I would accept it! |
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Mar
14 |
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comment |
Image of Homomorphism of Lie groups Thank you, that is what I was looking for. At first, I thought that I must work the coordinate charts explicitely. I got the smoothness of multiplicaion myself. And for the kernel part, I can show that Lie group homomorphism must be of constant rank, and that means that level sets are embedded submanifolds of domain, and therefore closed. I still have to work out the quotienting of manifolds, but I know where to look now. |
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Mar
14 |
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Image of Homomorphism of Lie groups So it is really not clear to me how tu put topology and smooth structure on $Im(f)$ so that everything works out. |
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Mar
14 |
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Image of Homomorphism of Lie groups I am using definition which says that Lie subgroup is immersed submanifold (so not necesseraly with subspace topology). |
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Mar
13 |
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revised |
Image of Homomorphism of Lie groups added 4 characters in body |
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Mar
13 |
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asked | Image of Homomorphism of Lie groups |
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Feb
26 |
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comment |
Eilenberg Moore category On morphisms just Ff=Tf, no need for ordered pair, because morphism of T-algebras is just morphism in $C$ with some aditional properties. |
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Feb
13 |
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comment |
Does an adjoint pair fix a unit/counit pair? Yeah, but I think it is a fun exercise. |
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Feb
13 |
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revised |
Does an adjoint pair fix a unit/counit pair? edited body |
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Feb
13 |
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revised |
Does an adjoint pair fix a unit/counit pair? added 675 characters in body |
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Feb
13 |
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answered | Does an adjoint pair fix a unit/counit pair? |
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Jan
28 |
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comment |
Help with a proof of Zorn's lemma Dear Brian M. Scott, thank very much you for your help! |
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Jan
28 |
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accepted | Help with a proof of Zorn's lemma |
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Jan
28 |
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revised |
Help with a proof of Zorn's lemma added 124 characters in body |
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Jan
28 |
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comment |
Help with a proof of Zorn's lemma The argument seems circular. How do I know $X=\{u \in L \colon u<z\} \supseteq \{u∈L_1 \colon u<z\}$? And moreover, how do I see that $z=\inf_{L_1}(L_1\setminus L)$ exists? (I can't use that $L$ is initial segment in $L_1$ because I want to prove that.) |
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Jan
28 |
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Help with a proof of Zorn's lemma That would solve my problems if I could prove it! So if $x \in L$, $y \in L_1$ and $y<x$, i must see that $y \in L$. If $L_1 \subseteq L$ I am finished, so assume $L \subsetneq L_1$. How can I derive contradiction? I guess using the definition of $\cal L_0$ in some way, but i don't seem to manage it. |