Does there exist any non-contractible manifold with fixed point property? Does there exist any non-trivial space (i.e not deformation retract onto a point) in $\mathbb R^n$ such that any continuous map from the space onto itself has a fixed point. I highly suspect that the quasi circle on $\mathbb R^2$ is an example. Yet I've not written down the (dirty) proof. But in this case all its homotpy groups are trivial. So if I assume my space as a manifold, then (QUESTION:) does this fixed point property force it to become a contractible manifold? I read somewhere that there exists a contractible compact manifold which does not satisfy this fixed point property. So does there exist any non-contractible manifold (compact) where this property follows? Or otherwise can anyone please provide an outline of how to prove that such a manifold is contractible? 
 A: Another nice example is $\mathbb RP^{2n}$. Here is a simple proof...
If $f:\mathbb RP^{2n} \to \mathbb RP^{2n}$ is a continuous function then look at the lift of the map $\bar{f} :S^{2n}\to S^{2n}$. A fixed point of $f$ is equivalent to a point $x\in S^{2n}$ s.t. $\bar{f}(x)=x \ or \ -x$. If such a point doesn't exist, then we can have a homotopy between Identitity and Antipodal map via $H(x,t)=cos(t)x+sin(t)\bar{f}(x)$.[This is well defined map from sphere to sphere]. But for even dimensional sphere antipodal map has degree $-1$. That's a contradiction.
A: Take the space $\mathbb{CP}^2$. Its cohomology ring is given by $\mathbb{Z}[a]/a^3$, where $a$ has degree $2$. A map $f:\mathbb{CP}^2\rightarrow \mathbb{CP}^2$ induces a map on the second cohomology group with $f^*(a)=k a$ with $k\in \mathbb{Z}$. From this you can compute the action on (co)homology on the other degrees. In degree zero it is the identity and on the fourth degree it is given by multiplying with $k^2$. Then the Lefschetz number of this map is seen to be $L(f)=k^2+k+1$. This number is never zero. A non-zero Lefschetz number implies a fixed point by the Lefschetz fixed point Theorem. 
