Moving vertex without moving adjacent edges

Question

I draw this graph with Tikz but I don't like how the edges leave {e}:

It would be perfect, if I could move the vertex {e} a little bit to the left withuot moving its adjacent edges, so that it looks more like in vertex {a}. Any ideas?

\begin{tikzpicture}[every node/.style={rectangle,draw=black,thick}]
\node(a)at(-2,1){\(\{a\}\)};
\node(b1)at(0,0){\(\{b_0\}\)};
\node(b0)at(0,2){\(\{b_1\}\)};
\node(c1)at(2,0){\(\{c_0\}\)};
\node(c0)at(2,2){\(\{c_1\}\)};
\node(d1)at(5,0){\(\{d_3,\hdots,d_9\}\)};
\node(d0)at(5,2){\(\{d_0,\hdots,d_2\}\)};
\node(e)at(8,1){\(\{e\}\)};
\draw(a)to[bend left=30](b0);
\draw(a)to[bend right=30](b1);
\draw(e)to[bend right=20](d0);
\draw(e)to[bend left=20](d1);
\draw(b0)--(c0)--(d0)--(d1)--(c1)--(b1);
\end{tikzpicture}

Answer 1

It is not the node {e} that should be moved, it is the entrance of the edge in {e}. You can specify this by e.g.

\draw(e.north)to[bend right=20](d0);
\draw(e.south)to[bend left=20](d1);

which gives

You can also give the position as an angle, like in

\draw(e.90)to[bend right=20](d0);
\draw(e.-90)to[bend left=20](d1);

which gives the same picture.

If you also want to change the angle of which the edge is starting you can, apart from [bend right=<angle>] give different angles for in and out, as in

\draw(e.north)to[out=90,in=0](d0);
\draw(e.south)to[out=-90,in=0](d1);

In this case the positioning e.north and e.south does not make any difference, but with some different settings of the line angles it will. (Then it might also look better than this example.)