模空间的拉回图

Question

我问过这个问题之前。现在，我想进一步丰富一下这个图表。

我想用 egreg 的答案画回拉系列，但是上面的箭头看起来很奇怪。我只是修改了他的答案：

\begin{tikzcd}[row sep=3em,column sep=3em]
B'\mathrlap{{}\times\mathbb{P}^1}
\arrow[d,"\pi'"',"\;\bigg\uparrow\bigg\uparrow\bigg\uparrow\bigg\uparrow\sigma_i'"]\arrow[rr] 
& & B\mathrlap{{}\times\mathbb{P}^1}
\arrow[d,"\pi"',"\;\bigg\uparrow\bigg\uparrow\bigg\uparrow\bigg\uparrow\sigma_i"]\\
B'\arrow[rr,"\varphi"] & & B
\end{tikzcd}.

那么上面的箭头就是错误的……：

。

Answer 1

对你的问题的一个可能的答案是：使用shorten。这既适用于egreg的代码，也适用于我对你上一个问题的回答。

\documentclass{article}
\usepackage{tikz-cd}
\usepackage{mathtools,amssymb}
\begin{document}
\begin{tikzcd}[row sep=3em,column sep=3em]
B'\mathrlap{{}\times\mathbb{P}^1}
\arrow[d,"\pi'"',"\;\bigg\uparrow\bigg\uparrow\bigg\uparrow\bigg\uparrow\sigma_i'"]
\arrow[rr,shorten <=2.1em] 
& & B\mathrlap{{}\times\mathbb{P}^1}
\arrow[d,"\pi"',"\;\bigg\uparrow\bigg\uparrow\bigg\uparrow\bigg\uparrow\sigma_i"]\\
B'\arrow[rr,"\varphi"] & & B
\end{tikzcd}.


\begin{tikzcd}[column sep=4.5em,row sep=2.5em,execute at end picture={
\foreach \Y in {1,2} {\foreach \X in {1,2,3,4}
{\draw[latex-,shorten >=1pt,shorten <=1pt] ([xshift=\X*1ex-1ex]M1\Y.south east)  coordinate
(aux-\X) -- (aux-\X|-M2\Y.north)
\ifnum\X=4
node[midway,right] {$\sigma_i\ifnum\Y=1 '\fi$}
\fi;}}
}]
 |[alias=M11,text width=width("$B'$")]|B'\times \mathbb{P}^1  
 \arrow[r,shorten <=2.1em] \arrow[d,"\pi'" swap]
 &  |[alias=M12,text width=width("$B$")]|B\times \mathbb{P}^1\arrow[d,"\pi" swap]  \\
 |[alias=M21]| B' \arrow[r,"\phi" swap]  & |[alias=M22]| B \\
\end{tikzcd}
\end{document}