Re: [NTG-context] I can't put footnote

22 Jul 2009

      En/na Hans Hagen ha escrit:
...
Xan wrote:
...
No, with the version line commented I get the same error:
i don't get that error so i need a smaller test file
Hans
I hope this serves you. Now I have another error. What fails???

Xan.

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\subsubsubject{El problema de la paraula}

Sigui $G$ un grup i $X$ un subconjunt de $G$. Aleshores l'aplicació $\pi \colon {(X \cup X^{-1})}^* \to G$\symbol{$\pi$} consistent a enviar cada lletra $x \in X$ a l'element corresponent $\pi(x) = x \in G$ es pot estendre de manera natural a totes les paraules de $X \cup X^{-1}$ de la forma següent:
\startitemize[n]
\item $\pi(x^{-1}) = \pi(x)^{-1}$ per a tot $x \in X$.
\item Per a tota paraula $w = w_1 \ldots w_r$ sobre $X \cup X^{-1}$,
\startformula
\pi(w) = \pi(w_1) \cdots \pi(w_r) \in G.
\stopformula
\stopitemize
Per tant, $\pi$ és un morfisme de monoides entre ${(X \cup X^{-1})}^*$ i $G$. De forma òbvia, si $X$ és un conjunt de generadors de $G$, aleshores $\pi$ és exhaustiva. En particular, si ${\cal P} = \langle X \mid R \rangle$ és una presentació, aleshores, com que $X$ és un conjunt de generadors de $G({\cal P})%$\footnote{Recordem que abusem del llenguatge, identificant $\iota(X)$ i $X$, i que, realment, $\iota(X)$ és el generador de $G(\cal{P})$.}
, llavors $\pi \colon {(X \cup X^{-1})}^* \to G({\cal P})$ és un morfisme exhaustiu.

\stoptext

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subsubsubject   : - El problema de la paraula
! Missing $ inserted.
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                $
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l.245 , llavors $\pi \colon {(X \cup X^
                                       {-1})}^* \to G({\cal P})$ és un morf...

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