Does a self map from the wedge sum of two spheres have either a fixed point or a point of period 2? Let $X$ be the wedge sum of two $2$-dimensional spheres and $f$ a continuous function from $X$ into $X$. Does $f$ have either a fixed point or a periodic point of order 2?
Thanks
 A: Will's comment above shows that the action of $f$ on $H^2$ is of order 3
(the eigenvalues are the nontrivial third roots of unity) and something like
$\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$. So I thought 
one should try a construction using third roots of unity realizing this.
Write the wedge as $S^1 \times [0,2] / \sim $ where one identifies
$(s,0)\sim (s',0),  (s,1)\sim (s',1), (s,2)\sim (s',2)$.
Let $\zeta$ be a third root of unity, and
define 
$f: S^1 \times [0,2] / \sim\  \to \ S^1 \times [0,2] / \sim$
by 
$f(s,t)=(\zeta s, 2-2t)$ for $t\le 1$ and $f(s,t)=(\zeta s, t-1)$ for $t\ge 1$
Note that the three points where $t=0,t=1,t=2$ are permuted cyclically.
This implies $f$ and $f^2$ have no fixed points (for all other points the $\zeta$ respectively $\zeta^2$ factor works).
Note also that multiplication by $\zeta$ is not really necessary, almost any rotation of the circle would be fine. Multiplication by $\zeta$ allows $f^3$ to be close to the identity. If one uses a different rotation, the three points with $t=0,t=1,t=2$ are the only fixed points. 
