Any solutions ?

I have moved this from the Algebra subforum to Topology. I hope it is right here. 1) Not CN. Simply take the map which sends every point on the sphere to its antipode. 2) CN. Let $C$ be this cross, and let $f:C\to C$ be a continuous map. Assume (for the sake of contradiction) that $f$ has no fixed point. Let $\pi_x:\mathbb R^2\to\mathbb R$ be the map sending every point to its $y$-coordinate. Consider the function $\mathbb R\to \mathbb R,\ t\mapsto\pi_x\left(f\left(t,0\right)\right)-t$. This function is clearly continuous. Its value at $t=-1$ is nonnegative (because $\pi_x\left(f\left(-1,0\right)\right)$ must be $\geq -1$). Its value at $t=1$ is nonpositive (for similar reasons). Thus, by the intermediate value theorem, it must take the value $0$ at $t=t_0$ for some $t_0\in\left[-1,1\right]$. If this $t_0$ is not zero, then $\left(t_0,0\right)$ must then be a fixed point of $f$ (because the only way we can have $\pi_x\left(f\left(t_0,0\right)\right)-t_0=0$ with $t_0\neq 0$ is when $f\left(t_0,0\right)=\left(t_0,0\right)$), contradicting our assumption that $f$ have no fixed points. Thus, $t_0$ must be zero. Then, $\pi_x\left(f\left(0,0\right)\right)-0=0$. In other words, $\pi_x\left(f\left(0,0\right)\right)=0$. So the $x$-coordinate of $f\left(0,0\right)$ must be $0$. Similarly, the $y$-coordinate of $f\left(0,0\right)$ must also be $0$. Thus, $f\left(0,0\right)=\left(0,0\right)$, again contradicting the assumption that $f$ have no fixed point. This contradiction proves our assumption wrong, and thus the cross $C$ is CN. 3) CN. Let $G$ denote the graph of $f$. Let $h:G\to G$ be a continuous map. Assume (for the sake of contradiction) that $h$ has no fixed point. Let $\pi_x:\mathbb R^2\to\mathbb R$ denote the map which takes every point to its $x$-coordinate. Clearly, $\pi_x\mid _G$ is injective and continuous. Let me rename the function $f$ as $p$, and use the notation $f$ for the function $\left[0,1\right] \to G,\ x\mapsto \left(x,p\left(x\right)\right)$. (Sorry for this renaming, but I am too tired to fix what otherwise would be a lot of mistakes...) For every $t\in\left[0,1\right]$, we notice that $f\left(t\right)$ is the unique point on $G$ whose $x$-coordinate is $t$. Thus, the map $f:\left[0,1\right]\to G$ is bijective (being the inverse of $\pi_x\mid _G$) and continuous on $\left(0,1\right]$ (but not on $\left[0,1\right]$). Clearly, $\pi_x\left(h\left(f\left(1\right)\right)\right)\leq 1=\pi_x\left(f\left(1\right)\right)$. Hence, if we can find some $t>0$ satisfying $\pi_x\left(h\left(f\left(t\right)\right)\right)\geq \pi_x\left(f\left(t\right)\right)$, then by the intermediate value theorem we can find some $t_0>0$ satisfying $\pi_x\left(h\left(f\left(t_0\right)\right)\right)= \pi_x\left(f\left(t_0\right)\right)$, which yields that $f\left(t_0\right)$ is a fixed point under $h$ (since $\pi_x\mid _G$ is injective), contradicting the assumption that $h$ have no fixed point. Thus, we must have $\pi_x\left(h\left(f\left(t\right)\right)\right) < \pi_x\left(f\left(t\right)\right)$ for every $t>0$. Since $h$ has no fixed point, we have $h\left(f\left(0\right)\right)\neq f\left(0\right)$. Thus, $h\left(f\left(0\right)\right) = f\left(u\right)$ for some $u\in\left(0,1\right]$. Consider this $u$. Then, $u>0$, so that $\pi_x\left(u\right)>0$. Now let $V$ be an open neighbourhood of $f\left(u\right)$ in $G$ such that $\pi_x\left(V\right)$ is bounded away from $0$ (such a $V$ obviously exists since $\pi_x\left(u\right)>0$). Let $U=h^{-1}\left(V\right)$. Since $h$ is continuous, it follows that $U$ is open, and clearly $f\left(0\right)\in U$ (since $h\left(f\left(0\right)\right)=f\left(u\right)\in V$). Since $\pi_x\left(V\right)$ is bounded away from $0$, there exists some $\epsilon>0$ such that $B\left(0,\epsilon\right) < \pi_x\left(V\right)$ (this means that every element of $B\left(0,\epsilon\right)$ is smaller than every element of $\pi_x\left(V\right)$). Consider this $\epsilon$. Let $U_1=U\cap \pi_x^{-1}\left(B\left(0,\epsilon\right)\right)$. Then, $U_1$ is an open subset of $U$ still containing $f\left(0\right)$ but satisfying $\pi_x\left(U_1\right) < \pi_x\left(V\right)$. Since the topology on $G$ was defined as the restriction of the Euclidean topology on $\mathbb R^2$, it is easy to see that there exists some $q\in \left(0,1\right]$ such that $f\left(q\right)\in U_1$ (in fact, we can choose $q$ as $\frac{1}{\pi N}$ for a sufficiently high $N\in\mathbb N$). Consider this $q$. Then, $f\left(q\right)\in U_1\subseteq U=h^{-1}\left(V\right)$, so that $h\left(f\left(q\right)\right)\in V$. Thus, $\pi_x\left(h\left(f\left(q\right)\right)\right) \in \pi_x\left(V\right)$. Combined with $\pi_x\left(f\left(q\right)\right) \in \pi_x\left(U_1\right)$ (since $f\left(q\right)\in U_1$), this yields $\pi_x\left(f\left(q\right)\right) < \pi_x\left(h\left(f\left(q\right)\right)\right)$ (since $\pi_x\left(U_1\right) < \pi_x\left(V\right)$). But this contradicts the fact that $\pi_x\left(h\left(f\left(t\right)\right)\right) < \pi_x\left(f\left(t\right)\right)$ for every $t>0$. This shows that our assumption was wrong, and $h$ must have a fixed point, qed. For similar results, see R. H. Bing, The Elusive Fixed Point Property, The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 119-132.