  The problem Simultaneous Embedding with Fixed Edges (SEFE) asks for two
planar graph $G^1 = (V^1, E^1)$ and $G^2 = (V^2, E^2)$ sharing a common
subgraph $G = G^1 \cap G^2$ whether they admit planar drawings such that the
common graph is drawn the same in both. Previous results on this problem
require $G$, $G^1$ and $G^2$ to be connected. This paper is a first step
towards solving instances where these graphs are disconnected.
  First, we show that an instance of the general SEFE-problem can be reduced in
linear time to an equivalent instance where $V^1 = V^2$ and $G^1$ and $G^2$ are
connected. This shows that it can be assumed without loss of generality that
both input graphs are connected. Second, we consider instances where $G$ is
disconnected. We show that SEFE can be solved in linear time if $G$ is a family
of disjoint cycles by introducing the CC-tree, which represents all
simultaneous embeddings. We extend these results (including the CC-tree) to the
case where $G$ consists of arbitrary connected components, each with a fixed
embedding.
  Note that previous results require $G$ to be connected and thus do not need
to care about relative positions of connected components. By contrast, we
assume the embedding of each connected component to be fixed and thus focus on
these relative positions. As SEFE requires to deal with both, embeddings of
connected components and their relative positions, this complements previous
work.
