将交换图纳入问答框架的问题

Question

当我通过 TeXLive 2022 运行您的代码时，它会产生多个错误。这是因为您显然将相同的代码粘贴了environ两次subproof。我会帮您删除它。

无论哪种方式，您都需要使用ampersand replacement选项因为Environ环境将抓住环境的主体，此时 TikZ 没有机会再改变 catcode &（在这种情况下它是一种假对齐）。

通常使用\&就可以了（因为该&字符在矩阵中不怎么使用）：

\begin{tikzcd}[row sep=huge, ampersand replacement=\&]
      \& X \ar[dl,"id",swap,sloped] \ar[dr,"\rho_{B}",sloped] \ar[d,dashed,"{u}" description] \\
    X \& X \times \textbf{1}\ar[l,"\pi_{X}"] \ar[r,"\pi_{\textbf{1}}",swap] \& \textbf{1}
\end{tikzcd}

但是，由于所有 TikZ-CD 图（实际上是所有 TikZ 矩阵）都problem需要这个，我建议添加

\tikzset{ampersand replacement=\&}

到problem环境中。
您仍然需要在图表中使用\&而不是，&但您不必再指定ampersand replacement=\&。

无论是环境\[\]还是示例都无法很好地处理显示数学内容，这就是我在这里使用环境来使图表居中的原因。（TikZ-CD 确保内容无论如何都处于数学模式，因此无论如何都没有技术理由在这里使用数学环境 - 除非您想要方程编号。）equation*amsmathtabularxcenter

不要手动使用(a)、(b)等来标记，我已经添加了，\enumalphparen以便enumerate在环境内部可以使用problem标记\items (a)、、(b)...
```
\newcommand*\enumalphparen{%
  \renewcommand*\labelenumi{\theenumi}%
  \renewcommand*\theenumi{(\alph{enumi})}}%
```
使用该enumerate环境还可以让您方便地使用\label-\ref系统。
不要\textbf在数学模式中用于数学内容。请\mathbf改用。
不要使用\mathbf。使用逻辑语法。我已经定义了
```
 \newcommand*\obj[1]{\mathbf{#1}}
```
这样你就不必\obj{1}考虑对象是如何排版的了。
解决方案部分中的前面enumerate没有段落。使用\vspace{-\topskip}将第一项拉到同一高度。

代码

\documentclass{amsart}
\usepackage{latexsym,amsmath}
\usepackage{tabularx, environ}
\usepackage{wasysym,latexsym,amsmath,amssymb,mathrsfs}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{mathtools}
\newcommand{\one}{\obj{1}}

\usepackage{scalerel,stackengine}
\usepackage{tikz-cd}
\usepackage{pst-node}
\DeclareMathOperator{\id}{id}
\stackMath
\newcommand\reallywidehat[1]{%
\savestack{\tmpbox}{\stretchto{%
  \scaleto{%
    \scalerel*[\widthof{\ensuremath{#1}}]{\kern-.6pt\bigwedge\kern-.6pt}%
    {\rule[-\textheight/2]{1ex}{\textheight}}%WIDTH-LIMITED BIG WEDGE
  }{\textheight}% 
}{0.5ex}}%
\stackon[1pt]{#1}{\tmpbox}%
}
\parskip 1ex

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}[theorem]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}{Definition}[theorem]
\newtheorem{remark}{Remark}[theorem]
\newtheorem{question}{Question}[theorem]
\newtheorem{solution}{Solution}[theorem]
\makeatletter

% https://tex.stackexchange.com/a/199244/26355
\newcolumntype{\expand}{}
\long\@namedef{NC@rewrite@\string\expand}{\expandafter\NC@find}
\NewEnviron{problem}[2][]{%
  \tikzset{ampersand replacement=\&}% ← added !!
  \enumalphparen                    % ← added !!
  \def\problem@arg{#1}%
  \def\problem@framed{framed}%
  \def\problem@lined{lined}%
  \def\problem@doublelined{doublelined}%
  \ifx\problem@arg\@empty%
    \def\problem@hline{}%
  \else%
    \ifx\problem@arg\problem@doublelined%
      \def\problem@hline{\hline\hline}%
    \else%
      \def\problem@hline{\hline}%
    \fi%
  \fi%
  \ifx\problem@arg\problem@framed%
    \def\problem@tablelayout{|>{\bfseries}lX|c}%
    \def\problem@title{\multicolumn{2}{|l|}{%
        \raisebox{-\fboxsep}{\textsc{\Large #2}}%
      }}%
  \else
    \def\problem@tablelayout{>{\bfseries}lXc}%
    \def\problem@title{\multicolumn{2}{l}{%
        \raisebox{-\fboxsep}{\textsc{\Large #2}}%
      }}%
  \fi%
  \bigskip\par\noindent%
  \renewcommand{\arraystretch}{1.2}%
  \begin{tabularx}{\textwidth}{\expand\problem@tablelayout}%
    \problem@hline%
    \problem@title\\[2\fboxsep]%
    \BODY\\\problem@hline%
  \end{tabularx}%
  \medskip\par%
}
\makeatother
\newenvironment{subproof}[1][\proofname]{%
  \renewcommand{\qedsymbol}{$\blacksquare$}%
  \begin{proof}[#1]%
}{%
  \end{proof}%
}
\usepackage{tikz-cd}
%%% Added:
\newcommand*\enumalphparen{%
  \renewcommand*\labelenumi{\theenumi}%
  \renewcommand*\theenumi{(\alph{enumi})}}%
\newcommand*\obj[1]{\mathbf{#1}}
%%%

\begin{document}
\begin{problem}[framed]{Question 2}
  %Task: & Define a new ``problem'' environment. \\
  Problem: & Show that any two terminal objects are isomorphic. Specifically, if $\obj{1}$ and $\obj{1}^{\prime}$ are both terminal objects, show that there are arrows $f: \obj{1} \rightarrow \obj{1}^{\prime}$ and $g: \obj{1}^{\prime} \rightarrow \obj{1}$ such that $g \circ f=\mathrm{id}_{1}$ and $f \circ g=\mathrm{id}_{1}$. This is what it means for two objects to be isomorphic in a category.\\
  Solution: & As, $\obj{1},\obj{1}^{\prime}$ are terminal objects. From $\obj{1} \to \obj{1}^{\prime}, \obj{1} \to \obj{1}$ we have a unique map so are  from $\obj{1}^{\prime} \to \obj{1}, \obj{1}^{\prime}\to \obj{1}^{\prime}$. Then we have maps $f \circ g: \obj{1}^{\prime} \to \obj{1}^{\prime}$ and $g \circ f: \obj{1} \to \obj{1}.$ Now $id|_{\obj{1}^{\prime}}:\obj{1}^{\prime} \to \obj{1}^{\prime}$ and $id|_{\obj{1}}:\obj{1} \to \obj{1}$ are also maps then $g \circ f=id_{\obj{1}}$ and $f \circ g=id_{\obj{1}^{\prime}}.$
\end{problem}
 
\begin{problem}[framed]{Question 5}
  %Task: & Define a new ``problem'' environment. \\
  Problem: &
    Let $\obj{C}$ be a category with terminal object $\obj{1}$.
    Let $X$ be an object of $\obj{C}$.
    We will prove that $X \times 1$ is isomorphic to $X$.
\begin{enumerate}
  \item Because $X \times 1$ is the product of $X$ and 1,
         it comes with two arrows.
        Describe them by stating their domain and codomains.
  \item There is another object with arrows to both $X$ and 1.
         What is it? What are the arrows?
  \item Use the existence property of products to find a right-inverse $u$
        for the arrow $\pi_{X}: X \times \obj{1} \rightarrow X$. \label{item:c}
  \item Now we want to show that $u$ is a left-inverse to $\pi_{X}$ as well.
        To that end, name one arrow
          $f: X \times \obj{1} \rightarrow X \times \obj{1}$
        so that the following diagram commutes,
        meaning that $\pi_{X} \circ f=\pi_{X}$. \label{item:d}
    \begin{center}
    \begin{tikzcd}[row sep=huge]
      \& X\times \obj{1} \ar[dl,"\pi_X", swap, sloped]
                         \ar[dr]
                         \ar[d,"f" description] \\
    X \& X\times \obj{1} \ar[l,"\pi_X"]
                        \ar[r,"",swap]
      \& \obj{1}
    \end{tikzcd}
    \end{center}
  \item There is another arrow $X \times 1 \rightarrow X \times 1$
        so that the above diagram commutes.
        What is it?
        (Hint: it's a composition of two other arrows.) \label{item:e}
  \item Use the uniqueness property of products to conclude
        that the two arrows from items \ref{item:d} and
        \ref{item:e} must be equal. 
  \item Conclude that the arrow $u$ from part \ref{item:c}
        is a left-inverse to
          $\pi_{X}: X \times 1 \rightarrow X$
        as well as a right inverse.
  \item Conclude that $u$ is an isomorphism, and $X \cong X \times 1$.
\end{enumerate}\\
%%%%%
Solution: & \vspace{-\topskip} % no text here, remove topskip spacing from enumerate
\begin{enumerate}
  \item Domain and codomain of $\pi_X$ are $ X\times \obj{1}$ and $X$ resp.
        Domain and codomain of $\pi_{\obj{1}}$
        are $ X\times \obj{1}$ and $\obj{1}$ resp.
  \item Another object is $X$ and the arrows $X \to X$ is identity
        and $X \to \obj{1}$ is the unique map
        corresponding to the terminal object.
  \item If we replace $A$ by $X$ and $B$ by
        $\obj{1}, \rho_A=id, \pi_A=\pi_X$ in the figure of definition 2, 
    \begin{center}
    \begin{tikzcd}[row sep=huge]
          \& X \ar[dl, "id", swap, sloped]
               \ar[dr, "\rho_{B}", sloped]
               \ar[d, dashed, "u" description] \\
        X \& X \times \obj{1}
               \ar[l, "\pi_X"]
               \ar[r, "\pi_{\obj{1}}", swap]
          \& \obj{1}
    \end{tikzcd}
    \end{center}
        we'll get the unique map from $u:X \to X \times \obj{1}$
        s.\,t. $\pi_X \circ u=id_X$.
        Hence we get the right-inverse $\pi_A$ of $u$.
        Observe that $u: X \to X \times \obj{1}$ is the inclusion map here
        i.\,e. $u(x)=\{x\}\times \obj{1}.$
   \item %We can see that $u \circ \pi_X(\{x\} \times \obj{1})=u(x)=\{x\} \times \obj{1}$, i.e., $$X \times \obj{1} \stackrel{\pi_X}{\longrightarrow} X  \stackrel{u}{\longrightarrow} X \times \obj{1}
%$$ is the identity map. Hence $u$ is also a left-inverse of $\pi_X$. 
         We name $f=id_{X \times \obj{1}}.$
  \item $u \circ \pi_X$ is another arrow $X \times 1 \rightarrow X \times 1$
        so that
        $\pi_x \circ (u \circ \pi_X)=(\pi_x \circ u) \circ \pi_X=(id) \circ \pi_X=\pi_X$
        hence, the above diagram commutes.
  \item From Universal property of product we get $u \circ \pi_X=id_{X \times \obj{1}}.$
  \item $u$ is a left inverse of $\pi_X.$
  \item $X \cong X \times \obj{1}.$
\end{enumerate}
\end{problem}

\end{document}

输出

Answer 1