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Este es un teorema que establece un resultado acerca de productos libres y productos libres amalgamados de grupoides.


THM AEditar

Teorema de Ordman

Mapa de resultados del artículo del 1971: "On subgroups of amalgamated free products"

1 EnunciatoEditar

Thm A:Editar

Let

$ \bullet $ $ G=*_{G_0}G_{\mu\in M} $,
$ \bullet $ $ K=*_{\mu\in M}K_{\mu} $,
$ \bullet $ $ f:G\to K $ a homomorphism such that $ fG_{\mu}\subset K_{\mu} $

be

then

A(5)Editar

For each $ H < G $ we have $ H=*_{H_0}H_{\mu\in M} $ with $ fH_{\mu}\subset K_{\mu} $

A(6)Editar

$ H_0=\mathrm{gen}\{g_{0\nu}G_{0\nu}g_{0\nu}^{-1} :\ G_{0\nu} < G_0,\ g_{0\nu}\in{\mathrm{ker}}f \} $

A(7)Editar

For each $ H_{\mu} $ is generated by $ \{g_{\mu\nu}G_{\mu\nu}g_{\mu\nu}^{-1}:\ G_{\mu\nu}\subset G_{\mu}\{g_{\mu\nu}\}\subset \{g_{0\nu}\}\subset {\mathrm{ker}}f \}\cup \{g_1g_2g_3:\ g_2\in G, g_1,g_3\in\{g_{0\nu}\}\} $

A(8)Editar

When $ G_0=\{1\} $ we have $ H_0=\{1\} $





Thm HigginsEditar

Let $ G = \ast_{\mu \in M} G_{\mu} $ and $ K = \ast_{\mu \in M}K_{\mu} $ free products. $ f: G \longrightarrow K $, $ fG_{\mu} \subset K_{\mu} $ $ \forall {\mu} $, $ H < G $ with $ fH = K $.

Then

$ H=\ast_{\mu \in M}H_{\mu} $ with $ fH_{\mu} \subset K_{\mu} $.

Thm GrushkoEditar

Let $ g:F \longrightarrow \ast_{\mu \in M}K_{\mu} $, $ F $ free group, $ g $ onto. Then $ F = \ast_{\mu \in M}F_{\mu} $ with $ gF_{\mu} \subset K_{\mu} $.

Def + PrelimEditar

2.1. GroupoidsEditar

A groupoid is a small category in which each map has an inverse.

$ A $ is a set with a multiplication defined on some subset $ A \times A $, such that
$ \textbf{(G1)} $ $ xy $ and $ yz $ are defined iff $ (xy)z $ or $ x(yz) $ is.
$ \textbf{(G2)} $ $ \forall x\in A $, $ \exists e(x)\in A $ with $ e(x)x $ defined, such that $ e(x)y = y $ y $ ze(x) = z $ whenever these products are defined.
$ \textbf{(G3)} $ $ \forall x\in A $, $ \exists x^{-1}\in A $ with $ xx^{-1} = e(x) $, $ x^{-1}x = e(x^{-1}) $.

A subgroupoid $ B $ of $ A $ is a subset of $ A $ which is a groupoid under the induced multiplication.

2.2Editar

2.3Editar

2.3.1Editar

2.3.2Editar

2.3.3Editar

Some lemmasEditar

Lemma 3.1.1Editar

Lemma 3.1.2Editar

Lemma 3.2.1Editar

Lemma 3.2.2Editar

Proof of them AEditar

4.1Editar

4.2Editar

4.3Editar

4.4Editar

RemarksEditar

5.1Editar

5.2Editar

5.3Editar

5.4Editar


Particularizar para...Editar

Study:

How is for:

$ f:G_1*_{G_0}G_2\to K_1*K_2 $?

And for

$ f:\{G_1,G_2,G_3\}*_{G_0}\to K_1*K_2*K_3 $?


$ f:\{G_1,G_2,G_3,G_4\}*_{G_0}\to K_1*K_2*K_3*K_4 $