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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.

Particularizar para...Editar

Study:

How is for:

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

And for

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

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