\documentclass{article}
\usepackage[x11names,dvipsnames]{xcolor}
\usepackage{tkz-euclide}
\usepackage{pgf,tikz,pgfplots,tikz-3dplot}
\usepackage{background}
\backgroundsetup{
placement=center,
scale=5,
contents={Jason Bowens},
opacity=.2
}
\usepackage{imakeidx}
\makeindex
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{calligra}
\newcommand{\I}{\textup{\large\calligra i}\,}
\usepackage{mathtools} %\xhookrightarrow
% \usepackage{graphicx}
\usepackage[most]{tcolorbox}
\usepackage[margin=1in,top=.71in,bottom=1in,paperwidth=8.5in,paperheight=11in]{geometry}
\makeatletter\let\gb\@gobble\makeatother
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhead{} % clear all header fields
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\fancyfoot[R,RO]{\thepage} % page number in "outer" position of footer line
\fancyfoot[L,LO]{
{\tiny Property of Jason Bowens
for Ivan Danilenko \\ Math 141 class at the \\ University of California, Berkeley
} } % other info in "inner" position of footer line
\begin{document}
\setcounter{secnumdepth}{-1}
\begin{tcolorbox}[enhanced jigsaw, opacityback=.2,colback=white,breakable,enhanced]
Question: How far is a {\tt general immersion} from an {\tt inclusion} of a {\tt submanifold}?
The best case senario.
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick] (0.2-.1,0.2+.1) arc (135:315:.07cm);
\draw[thick,->] (-1+.2,1)--(1-.2,1);
\draw[thick,->] (-1+.1,1-.1)--(0-.2,0+.2);
\draw[thick,->] (0+.2,0+.2)--(1-.2,1-.2);
\draw[](0,0)node[]{${\I}\gb{ii}(\mathcal{M}\gb{ii})$\footnote{{\tt submanifold} in $\mathcal{N}$}};
\draw[](-1,1)node[]{$\mathcal{M}$};
\draw[](1,1)node[]{$\mathcal{N}$};
\draw[](0,1+.1)node[]{${\I}$};
\draw[](0,1-.1)node[]{\tt immersion};
\draw[thick](-.6-.1,.6-.1)node[rotate=-45]{\tt dffmrphsm};
\draw[](.6+.1,.6-.1)node[rotate=45]{\tt inclusion};
\end{tikzpicture}
\end{center}
But we dont know what {\tt dffmrphsm} is?
This is not always the case.
\underline{Reason}: ${\I}$ may fail to be {\tt injecttive}
\begin{center}
\begin{tikzpicture}
\def\a{-3}\def\b{0.0}
\draw[] (\a,\b) circle (1cm);
\draw[blue] (\a-1,\b) node[]{$\bullet$};
\draw[] (\a+1,\b) node[red]{$\bullet$};
\draw[thick,->] (\a+1+.2,\b)--(\a+2-.2,\b);
\def\SIN{{cos((\x)r)/(1+(sin((\x)r))^2)},{sin((\x)r)*cos((\x)r)/(1+(sin((\x)r))^2)}}
\draw[domain={0}:{2*pi},samples=500] plot(\SIN);
\draw[Purple2] (0,0) node[]{$\bullet$};
\end{tikzpicture}
\end{center}
What the {\bf\color{red}} sees,
\\
\begin{center}
\begin{tikzpicture}[scale=1]
\draw[very thick, ->](-1+.2,0)--(-.3,0);
\def\k{-1.0}
\def\seg{{\k*sqrt(1-(\x)^2)},\x}
\draw[domain=-.25:.25] plot(\seg);
\draw[] (\k,0) node[blue]{$\bullet$};
\def\SIN{{cos((\x)r)/(1+(sin((\x)r))^2)},{sin((\x)r)*cos((\x)r)/(1+(sin((\x)r))^2)}}
\draw[domain={.35*pi}:{.65*pi},samples=500] plot(\SIN);
\draw[blue] (0,0) node[]{$\bullet$};
\end{tikzpicture}
\hspace{2cm}
\begin{tikzpicture}[scale=1]
\draw[very thick, ->](-1+.2,0)--(-.3,0);
\def\k{1.0}
\def\seg{{\k*sqrt(1-(\x)^2)-2},\x}
\draw[domain={-.25}:{.25}] plot(\seg);
\draw[] (\k-2,0) node[red]{$\bullet$};
\def\SIN{{cos((\x)r)/(1+(sin((\x)r))^2)},{sin((\x)r)*cos((\x)r)/(1+(sin((\x)r))^2)}}
\draw[domain={1.35*pi}:{1.65*pi},samples=500] plot(\SIN);
\draw[red] (0,0) node[]{$\bullet$};
\end{tikzpicture}
\end{center}
The {\tt image} is not a {\tt manifold}.
\\ \\
In contrast to {\tt local diffeomorphism}, requiring an {\tt immersion} to be {\tt bijective} doesn't solf the issue.
\\ \\
\underline{Reason 2}
\\
\begin{tikzpicture}
\draw[](-3,0)node[]{(}--(-2,0)node[]{)};
\draw[->,very thick] (-1.8,0)--(-1.2,0);
\def\a{0.0}\def\b{1.5*pi}\def\d{.1}
\draw[blue,domain=\a:\b,samples=500] plot({cos((\x)r)},{sin((\x)r)});
\draw[blue] ({cos((\b)r)},{sin((\b)r)})--(2,{sin((\b)r)});
\draw[blue] ({cos((\a)r)},{sin((\a)r)})--({cos((\a)r)},{sin((\b)r)+\d});
\draw[dashed] ({cos((\a)r)},{sin((\b)r)+\d/2}) circle (.15cm);
\end{tikzpicture}
\\
\begin{tikzpicture}
\draw[red] (0.6,-1)node[rotate=90]{(}--(0.6,1)node[rotate=90]{)};
\draw[blue,domain=0.7:2.2,samples=500] plot(\x,{sin(40*(1/\x)r)})node[right]{$\sin\left(\frac{1}{x}\right)$};
\draw[](-4,0)node[]{(}--(-3,0)node[]{)};
\draw[](-2,0)node[]{(}--(-1,0)node[]{)};
\draw[] (-3.5,0+.3) node[](1){};
\draw[] (-1.5,0-.3) node[](2){};
\draw[] (0.6-.1,0-.1) node[](3){};
\draw[] (1.5,-1-.2) node[](4){};
\draw[->] (1) [out=50,in=120] to (3);
\draw[->] (2) [out=-50,in=-120] to (4);
\end{tikzpicture}
\end{tcolorbox}
\end{document}
生产
我无法透过盒子看到我的水印。
而且，如果我想让背景有颜色 colback=red!20!并且想看透它，该怎么办？
我试过了enhanced jigsaw, opacityback=.2,...没用
答案1
如果您查看tcolorbox文档中的“绘图方案”部分，您将看到frame周围的tcolorbox不是框架，而是一个填充的矩形，稍后会被部分覆盖interior。因此，如果您想构建一个透明的框架tcolorbox，则需要更改默认frame和interior构造。这就是以下代码所做的。
\documentclass{article}
\usepackage{background}
\backgroundsetup{
placement=center,
scale=5,
contents={Jason Bowens},
opacity=1
}
\usepackage{lipsum}
\usepackage[most]{tcolorbox}
\begin{document}
\begin{tcolorbox}[%
enhanced,
% The default filled rectangle is just a frame now
frame code={\draw[rounded corners, blue, thick](frame.north west) rectangle (frame.south east);},
% No ìnterior` is drawn
interior empty,
]
\lipsum[1-2]
\end{tcolorbox}
\end{document}
如果需要背景颜色，可以使用opacityback选项。
您也可以忘记frame code解决方案，而使用opacityframe选项。但在这种情况下，请记住两种颜色将在文本背景上混合，如下面的第二个框所示。
\documentclass{article}
\usepackage{background}
\backgroundsetup{
placement=center,
scale=5,
contents={Jason Bowens},
opacity=1
}
\usepackage{lipsum}
\usepackage[most]{tcolorbox}
\begin{document}
\begin{tcolorbox}[%
enhanced,
% The default filled rectangle is just a frame now
frame code={\draw[rounded corners, blue, line width=.5mm]([shift={(.25mm,-.25mm)}]frame.north west) rectangle ([shift={(-.25mm,.25mm)}]frame.south east);},
colback=red!30, opacityback=.5
]
\lipsum[1-2]
\end{tcolorbox}
\begin{tcolorbox}[%
colframe=blue, opacityframe=.5,
colback=red!30, opacityback=.5
]
\lipsum[1-2]
\end{tcolorbox}
\end{document}
