Is a superposition of (anti)symmetric states (anti)symmetric? Let's say we have the following wavefunction of two identical particles, $A$ and $B$:
$$\frac{1}{2}[(\chi(A)\psi(B)\pm\psi(A)\chi(B))+(\phi(A)\eta(B)\pm\eta(A)\phi(B))]$$
Is this properly (anti)symmetric?
i.e. can it be put in the following form?
$$\frac{1}{\sqrt2}(f_1(A)f_2(B)\pm f_2(A)f_1(B))$$
 A: A superposition of (anti)symmetric states is always (anti)symmetric, but it is not necessarily decomposable. So, your

$$\frac{1}{2}[(\chi(A)\psi(B)\pm\psi(A)\chi(B))+(\phi(A)\eta(B)\pm\eta(A)\phi(B))]$$

can be put in the form
$$\frac{1}{\sqrt2}(f(A,B)\pm f(B,A)).$$
but in most cases there would exist no such functions $f_1$, $f_2$ that $f(A,B)=f_1(A)f_2(B)$.
A: Here's an elegant way to show that any linear combination of (anti)symmetric states is always (anti)symmetric.  We use Dirac notation here for the states, and we assume, for simplicity, that we are dealing with a two-component system so that states of the system are linear combinations of products $|\psi\rangle = |\psi_1\rangle|\psi_2\rangle$.
First, we define the exchange operator $P$ on products of states as the unique linear operator that "flips" factors of product states;
\begin{align}
    P|\psi_1\rangle|\psi_2\rangle = |\psi_2\rangle|\psi_1\rangle.
\end{align}
Second, recall that state is said to be (anti)symmetric provided when it is acted upon the by exchange operator $P$, it gets multiplied by a $\pm$ sign (+ for symmetric and - for antisymmetric);
\begin{align}
    P|\psi\rangle = \pm|\psi\rangle.
\end{align}
Now suppose that states $|\psi\rangle$ and $|\phi\rangle$ are both antisymmetric, then when we apply the exchange operator to any linear combination of them, we get
\begin{align}
    P(a|\psi\rangle+b|\phi\rangle) 
    &= aP|\psi\rangle + bP|\phi\rangle \\
    &= a(\pm|\psi\rangle)+b(\pm|\phi\rangle) \\
    &= \pm(a|\psi\rangle+b|\phi\rangle).
\end{align}
This shows that any linear combination of them has the property that the exchange operator acting on that linear combination gives the linear combination multiplied by a $\pm$ sign, so it's (anti)symmetric also!
Addendum. (Construction of the exchange operator)
Above, we defined the exchange operator as the unique linear operator whose action on (tensor) product states is to switch the factors.  That such an operator exists and is unique can be proven as follows.  
Let the total Hilbert space be $\mathcal H\otimes \mathcal H$.  Let $\{|n\rangle\}$ be a basis for $\mathcal H$, then the set $\{|n\rangle|m\rangle\}$ is a basis for $\mathcal H\otimes\mathcal H$, the so-called tensor product basis.  Recall that a linear operator is determined by its action on any basis.  Therefore, there is a unique linear operator $P$ on $\mathcal H\otimes\mathcal H$ satisfying the property
\begin{align}
    P|n\rangle|m\rangle = |m\rangle|n\rangle
\end{align}
for all pairs $(n,m)$.  It remains to show that $P$ switches the tensor factors for any product state $|a\rangle|b\rangle$.  This can be done by expanding the state in each factor using the basis we used to define $P$ as follows:
\begin{align}
    P|a\rangle|b\rangle 
    &= P\left(\sum_na_n|n\rangle\sum_m b_m|m\rangle\right)\\
    &= P\left(\sum_{nm}a_nb_m|n\rangle|m\rangle\right)\\
    &= \sum_{nm}a_nb_m P|n\rangle|m\rangle\\
    &= \sum_{nm}a_nb_m |m\rangle|n\rangle\\
    &=\sum_mb_m|m\rangle\sum_n a_n|n\rangle\\
    &= |b\rangle|a\rangle.
\end{align}
