help in finding errors

Question

I have written this code, but it does not come out as pdf. I need help in find errors or maybe the problem is in my acrobat reader. This is the code

\documentclass{article}
\usepackage[small,nohug,heads=littlevee]{diagrams}
\usepackage{amssymb,latexsym,amsmath}
\usepackage{array}
\usepackage{tikz}
\usetikzlibrary{matrix}
\usepackage{caption}
\begin{document}
\section{Van Kampen theorem}
\begin{theorem}
If \\ 
\vspace \begin{array}[c]{cccc}
A&  \stackrel{i}{\rightarrow} & B\\

\scriptstyle{j}{\downarrow }  \\

C    \\
\end{array}\\ is a data of groups $A$, $B$, and $C$ and homomorphisms $i$ and $j$. Then there exists a pushout $(P,j',i')$ where \begin{equation}P=\frac{B*C}{N} \end{equaition} and $j'(b)=bN,  i'(c)=cN$ where $N$ is the normal subgroup of $B*C$ generated by $\{i(a)j(a^{-1}): a\in A\}$. This pushout called {\bf amalgamated fee product}.
\end{theorem}
\begin{proof}if $a\in A$ then
\begin{eqnarray*}
  j'\circ i(a) &=& i(a)N \\
  &=& j(a) ({{i(a)}^{-1}j(a)})^{-1}\\
  &=& j(a)N \\
 &=&j'\circ j(a)\\
  \end{eqnarray*}
  hence $(P,j',i')$ is a solution.
  Now suppose that $(G,f,g)$ is another solution of the data. By the definition of the free product there exists unique homomorphism $\psi: B*C \rightarrow G$ with $\psi\mathop{\mid_{B}}=f$ and $\psi\mathop{\mid_{C}}=g$, if $b\in B$ and $c\in C$ then $\psi (bc)= f(b)g(c)$ and for all $a\in A$ we have $\psi (i(a)j(a^{-1}))= f\circ i(a). g\circ j(a^{-1}) =1$ because $f\circ i=g\circ j $. Hence $N\leq ker\psi $ \\ 
  Now define 
\begin{equation} \varphi: \frac {B*C}{N}\rightarrow G \end{equation} to be the homomorphism induced by $\psi $ with 
\begin{eqnarray*}
  \varphi \circ j'(b)&=&\varphi (bN)\\
  &=& \psi (b)= f(b)\\ 
\end{eqnarray*} 
and $\varphi \circ i'(c)= \varphi (cN)= \psi (c)=g(c)$ then the following diagram commutes\\

  \begin{figure}
 \begin{center}
\begin{tikzpicture}[x=1.00mm, y=1.00mm, inner xsep=0pt, inner ysep=0pt, outer xsep=0pt, outer ysep=0pt]
\path[line width=0mm] (24.26,37.44) rectangle +(79.52,59.18);
\draw(26.26,90.70) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $A$};
\draw(80.23,90.70) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $B$};
\draw(26.26,56.23) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $C$};
\draw(79.87,56.05) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $G$};
\definecolor{L}{rgb}{0,0,0}
\path[line width=0.30mm, draw=L] (30.78,91.79) -- (78.97,91.79);
\definecolor{F}{rgb}{0,0,0}
\path[line width=0.30mm, draw=L, fill=F] (78.97,91.79) -- (76.17,92.49) -- (78.27,91.79) -- (76.17,91.09) -- (78.97,91.79) -- cycle;
\path[line width=0.30mm, draw=L] (27.89,89.80) -- (27.89,61.82);
\path[line width=0.30mm, draw=L, fill=F] (27.89,61.82) -- (28.59,64.62) -- (27.89,62.52) -- (27.19,64.62) -- (27.89,61.82) -- cycle;
\path[line width=0.30mm, draw=L] (30.42,57.49) -- (78.07,57.49);
\path[line width=0.30mm, draw=L, fill=F] (78.07,57.49) -- (75.27,58.19) -- (77.37,57.49) -- (75.27,56.79) -- (78.07,57.49) -- cycle;
\path[line width=0.30mm, draw=L] (81.68,89.44) -- (81.86,61.10);
\path[line width=0.30mm, draw=L, fill=F] (81.86,61.10) -- (82.54,63.91) -- (81.85,61.80) -- (81.14,63.90) -- (81.86,61.10) -- cycle;
\draw(99.01,40.52) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $P$};
\path[line width=0.30mm, draw=L] (98.65,43.95) -- (84.03,56.77);
\path[line width=0.30mm, draw=L, fill=F] (84.03,56.77) -- (85.67,54.40) -- (84.55,56.31) -- (86.59,55.45) -- (84.03,56.77) -- cycle;
\path[line width=0.30mm, draw=L] (83.66,90.52) -- (99.73,45.76);
\path[line width=0.30mm, draw=L, fill=F] (99.73,45.76) -- (99.44,48.63) -- (99.49,46.42) -- (98.12,48.16) -- (99.73,45.76) -- cycle;
\path[line width=0.30mm, draw=L] (29.69,55.69) -- (98.10,41.97);
\path[line width=0.30mm, draw=L, fill=F] (98.10,41.97) -- (95.50,43.20) -- (97.42,42.11) -- (95.22,41.83) -- (98.10,41.97) -- cycle;
\draw(52.98,87.82) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $i$};
\draw(28.79,76.08) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $j$};
\draw(51.90,59.66) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $g$};
\draw(78.43,75.54) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $f$};
\draw(92.15,70.85) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $j'$};
\draw(59.66,44.68) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $i'$};
\draw(89.44,53.70) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont \varphi };
\end{tikzpicture}%
\caption{VAN_1}
 \end{center}
\end{figure} \\
It remains to show the uniqueness of \varphi. assume that {\varphi\mathop ^{~}}. Since the digram 
 \begin{figure}
\centering
\begin{tikzpicture}[x=1.00mm, y=1.00mm, inner xsep=0pt, inner ysep=0pt, outer xsep=0pt, outer ysep=0pt]
\path[line width=0mm] (38.16,42.32) rectangle +(51.62,49.43);
\draw(58.75,85.83) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $B$};
\draw(40.16,70.85) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $A$};
\draw(58.75,70.49) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $G$};
\draw(69.58,58.94) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $P$};
\draw(78.61,45.40) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont $B*C$};
\definecolor{L}{rgb}{0,0,0}
\path[line width=0.30mm, draw=L] (59.84,84.75) -- (59.84,75.18);
\definecolor{F}{rgb}{0,0,0}
\path[line width=0.30mm, draw=L, fill=F] (59.84,75.18) -- (60.54,77.98) -- (59.84,75.18) -- (59.14,77.98) -- (59.84,75.18) -- cycle;
\path[line width=0.30mm, draw=L] (44.13,72.11) -- (58.03,72.11);
\path[line width=0.30mm, draw=L, fill=F] (58.03,72.11) -- (55.23,72.81) -- (58.03,72.11) -- (55.23,71.41) -- (58.03,72.11) -- cycle;
\path[line width=0.30mm, draw=L] (80.96,50.45) -- (71.75,59.66);
\path[line width=0.30mm, draw=L, fill=F] (71.75,59.66) -- (73.24,57.18) -- (71.75,59.66) -- (74.23,58.17) -- (71.75,59.66) -- cycle;
\path[line width=0.30mm, draw=L] (69.22,62.91) -- (62.00,70.31);
\path[line width=0.30mm, draw=L, fill=F] (62.00,70.31) -- (63.46,67.81) -- (62.00,70.31) -- (64.46,68.79) -- (62.00,70.31) -- cycle;
\path[line width=0.30mm, draw=L] (61.46,84.93) -- (70.13,63.81);
\path[line width=0.30mm, draw=L, fill=F] (70.13,63.81) -- (69.71,66.66) -- (70.13,63.81) -- (68.42,66.13) -- (70.13,63.81) -- cycle;
\path[line width=0.30mm, draw=L] (42.33,70.31) -- (68.86,60.56);
\path[line width=0.30mm, draw=L, fill=F] (68.86,60.56) -- (66.48,62.18) -- (68.86,60.56) -- (65.99,60.87) -- (68.86,60.56) -- cycle;
\path[line width=0.30mm, draw=L] (62.55,88.18) .. controls (75.82,84.47) and (84.98,72.35) .. (84.93,58.57) .. controls (84.92,55.76) and (84.49,52.96) .. (83.66,50.27);
\path[line width=0.30mm, draw=L, fill=F] (83.66,50.27) -- (85.16,52.74) -- (83.66,50.27) -- (83.82,53.15) -- (83.66,50.27) -- cycle;
\path[line width=0.30mm, draw=L] (41.79,69.58) .. controls (47.38,57.70) and (58.57,49.43) .. (71.57,47.56) .. controls (73.72,47.25) and (75.90,47.13) .. (78.07,47.20);
\path[line width=0.30mm, draw=L, fill=F] (78.07,47.20) -- (75.25,47.81) -- (78.07,47.20) -- (75.29,46.41) -- (78.07,47.20) -- cycle;
\draw(62.18,65.07) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont \varphi\mathop ^{~}};
\draw(77.17,55.14) node[anchor=base west]{\fontsize{14.23}{17.07}\selectfont \nu };
\end{tikzpicture}%
\caption{VAN_2}
\end{figure} \\
\~{\varphi} \circ \nu =\psi 
\v

arphi \circ \nu =\varphi 
\nu  is subjective 
Given $p\inP$ choose $\omega \in B*C$, then $\nu (\omega ) = p$,so \~{\varphi}(p)=\~{\varphi}(\nu (\omega)) =\psi (\omega )= (\varphi \circ \nu )(\omega ) =\varphi (p) \Rightarrow \~{\varphi} =\varphi     
\en

d{proof} \\



\end{document}

Your file has dozens of errors that are reported by LaTeX so it is surprising that you get a pdf at all, Surely simple typos like ! LaTeX Error: \begin{equation} on input line 20 ended by \end{equaition}. are easy for you to fix before asking people for help? If there are any hard to understand errors make an example that just makes that one error and ask a question about that error. — David Carlisle, Apr 02 '13 at 20:53
see Common errors and The joy of TeX errors. Learning TeX and understanding .log file through errors is another part of life. — texenthusiast, Apr 02 '13 at 21:31

score 6 · Accepted Answer · edited Jun 09 '17 at 19:16

I've tried to amend the most evident LaTeX errors and also some of the language errors. I've also changed your \varphi\mathop^{~} into \tilde{\varphi} that seems what you wanted.

The biggest change, however is in the usage of tikz-cd for producing the commutative diagrams. Its syntax is quite easy to manage: the diagrams were produced in a few minutes (and my experience with the package is still limited).

\documentclass{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{array}
\usepackage{tikz,tikz-cd}

\newtheorem{theorem}{Theorem}

\begin{document}

\section{Van Kampen theorem}
\begin{theorem}
Let
\begin{equation*}
\begin{tikzcd}
A \arrow{r}{i} &  B \arrow{d}{j} \\
{} & C
\end{tikzcd}
\end{equation*}
be a data of groups $A$, $B$, and $C$ and homomorphisms $i$ and $j$. 
Then there exists a pushout $(P,j',i')$ where 
\begin{equation*}
P=\frac{B*C}{N}
\end{equation*}
and $j'(b)=bN$, $i'(c)=cN$ where $N$ is the normal subgroup of $B*C$ 
generated by $\{i(a)j(a^{-1}): a\in A\}$. This pushout is called 
\textbf{amalgamated free product}.
\end{theorem}

\begin{proof}
If $a\in A$ then
\begin{align*}
j'\circ i(a) &= i(a)N \\
 &= j(a) (i(a)^{-1}j(a))^{-1}\\
 &= j(a)N \\
 &= j'\circ j(a)
\end{align*}
hence $(P,j',i')$ is a solution.
Now suppose that $(G,f,g)$ is another solution of the data. By the 
definition of the free product there exists a unique homomorphism 
$\psi\colon B*C \rightarrow G$ with $\psi|_{B}=f$ and  $\psi|_{C}=g$. 
If $b\in B$ and $c\in C$, then $\psi (bc)= f(b)g(c)$ and for all 
$a\in A$ we have $\psi (i(a)j(a^{-1}))= f\circ i(a)$. $g\circ j(a^{-1}) =1$ 
because $f\circ i=g\circ j $. Hence $N\leq \ker\psi$.  

Now define 
\begin{equation*}
\varphi\colon \frac {B*C}{N}\rightarrow G
\end{equation*}
to be the homomorphism induced by $\psi $ with 
\begin{align*}
\varphi \circ j'(b)&=\varphi (bN)\\
  &= \psi (b)= f(b)
\end{align*} 
and $\varphi \circ i'(c)= \varphi (cN)= \psi (c)=g(c)$ 
then the following diagram commutes
\begin{equation*}
\begin{tikzcd}[column sep=3pc,row sep=2pc]
A \arrow{r}{i} \arrow{d}[swap]{j} &
B \arrow{d}[swap]{f} \arrow{ddr}{j'} \\
C \arrow{r}{g} \arrow{drr}[swap]{i'} &
G \\
{} & {} & P \arrow{lu}[swap]{\varphi}
\end{tikzcd}
\tag{VAN$_1$}
\end{equation*}
It remains to show the uniqueness of $\varphi$. Assume that $\tilde{\varphi}$ 
has the same property. Since the diagram 
\begin{equation*}
\begin{tikzcd}[column sep=2.5pc,row sep=2pc]
{} & B \arrow{d} \arrow[bend left]{dddrr} \\
A \arrow{r} \arrow{rrd} \arrow[bend right]{rrrdd} & G \\
{} & {} & P \arrow{ul}[swap]{\tilde{\varphi}} \\
{} & {} & {} & B*C \arrow{ul}[swap]{\nu}
\end{tikzcd}
\tag{VAN$_2$}
\end{equation*}
commutes, $ \tilde{\varphi}\circ \nu =\psi \varphi \circ \nu =\varphi $
and $\nu$  is surjective.

Given $p\in P$ choose $\omega\in B*C$, then $\nu (\omega ) = p$, so 
$\tilde{\varphi}(p)=\tilde{\varphi}(\nu (\omega)) =\psi (\omega )= 
(\varphi \circ \nu )(\omega ) =\varphi (p)$ implies  $\tilde{\varphi}=\varphi$
\end{proof}

\end{document}

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help in finding errors

1 Answers1

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