  It is known that every homeomorphism of the plane has a fixed point in a
non-separating, invariant subcontinuum. Easy examples show that a branched
covering map of the plane can be periodic point free. In this paper we show
that any branched covering map of the plane of degree with absolute value at
most two, which has an invariant, non-separating continuum $Y$, either has a
fixed point in $Y$, or $Y$ contains a \emph{minimal (by inclusion among
invariant continua), fully invariant, non-separating} subcontinuum $X$. In the
latter case, $f$ has to be of degree -2 and $X$ has exactly three fixed prime
ends, one corresponding to an \emph{outchannel} and the other two to
\emph{inchannels}.
