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我尝试修正最明显的 LaTeX 错误以及一些语言错误。我还将您的\varphi\mathop^{~}代码改成了\tilde{\varphi}您想要的样子。

然而，最大的变化在于使用tikz-cd来生成交换图。它的语法非常容易管理：几分钟内就可以生成交换图（而且我对这个软件包的经验仍然有限）。

\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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Answer 1

我尝试修正最明显的 LaTeX 错误以及一些语言错误。我还将您的\varphi\mathop^{~}代码改成了\tilde{\varphi}您想要的样子。

然而，最大的变化在于使用tikz-cd来生成交换图。它的语法非常容易管理：几分钟内就可以生成交换图（而且我对这个软件包的经验仍然有限）。

\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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帮助查找错误

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