# If $X$ has the fixed point property, what about $\text{Cont}(X,X)$?

If $$(X,\tau)$$ is a topological space, we denote by $$\text{Cont}(X,X)$$ the collection of all continous functions $$f:X\to X$$. We say that $$(X,\tau)$$ has the fixed point property if for any $$f\in\text{Cont}(X,X)$$ there is $$x_0\in X$$ such that $$f(x_0) = x_0$$.

If $$(X,\tau)$$ has the fixed point property, does $$\text{Cont}(X,X)$$ have the fixed point property?

• What topology do you put on Cont? I assumed the compact open topology. – Thomas Rot Nov 12 '18 at 14:51

No. In infinite dimensional normed spaces, there exists a retraction of the unit Ball onto the unit sphere. Since for $$X=[-1,1]$$ with the usual topology, $$C(X,X)$$ is simply the unit Ball of $$C(X)$$ and since the unit sphere of the latter admits a trivial fixed point free self map (multiplication with $$-1$$), we are done.
The question has already been answered. But here is a really simple argument. Take $$X=[a,b]$$ to be a closed interval and consider a map from $$C(X,X)$$ to itself that shifts the graph downwards in the first half of the interval and upwards in the second half. A fixed point has to be constantly $$a$$ in the first half of the interval and constantly $$b$$ in the second half, so can not be continuous.
For an explicit formula, take $$X=[-1,1]$$ and consider the map that takes $$f(x)$$ to $$(1-\vert x \vert) f(x) + x$$.
It suffices to find a space with the fixed point property such that $$\mathrm{Cont}(X,X)$$ has multiple connected components: Take a representative in each component and permute.
I think that $$\mathbb{CP}^2$$ has the fixed property and the space of self maps has multiple connected components. I hope I am not mistaken but I think that $$[\mathbb{CP}^2,\mathbb{CP}^2]\cong [\mathbb{CP}^2,\mathbb{CP}^\infty]\cong \mathbb{Z}$$. At least it is not too hard to find two self maps which are not homotopic: The constant map and the identity. Map all maps that are homotopic to the constant map to the identity and all maps that are not homotopic to the constant map to the constant map.
EDIT: I think you can prove very generally that $$\mathrm{Cont(X,X)}$$ does not have the fixed point property if $$X$$ is manifold of positive dimension. The space of continuous maps admits a model as a seperable Banach manifold. But any seperable Banach manifold of continuous maps is homeomorphic to a seperable Hilbert manifold $$X$$ (this is a theorem of Henderson (?)). Any separable infinite dimensional Hilbert manifold $$Y$$ is homeomorphic to $$Y\times \mathbb{R}$$. But for this you can write down a nonzero translation.