Transparent tcolorbox so I can see my watermark

Question

\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
\renewcommand{\headrulewidth}{0pt} % no line in header area
\fancyfoot{} % clear all footer fields
\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}

produces

I can't see my watermark through the box.

And also, what if I want to have color in my background colback=red!20! and I want to see through it?

I tried enhanced jigsaw, opacityback=.2,...didn't work

Answer 1

If you look at "Drawing Scheme" section in tcolorbox documentation, you will see that the frame around a tcolorbox is not a frame but a filled rectangle which is later on covered by the interior part. Therefore, if you want to build a transparent framed tcolorbox, you need to change the default frame and interior constructions. This is what is done in following code.

\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}

In case that a background color is desired, it's possible to use opacityback option.

You can also forget the frame code solution, and use the opacityframe option. But in this case remember that both colors will mix on the background of your text, as it's shown in following second box.

\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}