Fixed point property in topology I have a few questions concerning relating the fixed point property for a space $X$ (every continuous map from $X$ to $X$ has at least one fixed point) to some concepts in topology.
a). I know that a retract of a space with the fixed point property has a fixed point. However, if a space is a retract of another space, and the retract has the fixed point property, must the original space?
b). If two spaces $X$ and $Y$ have the same homotopy type, and $X$ has the fixed point property, need $Y$? 
c). Does the one point union of two spaces with the fixed point property also have it?
d). Does the unit ball with a pole removed have the fixed point property? How about if we require our map to be a homeomorphism?
I'm a little stuck relating topology to the fixed point concept, and I have my qualifying exam this Saturday, so having these questions answered would help. 
 A: a) If $X$ is an arbitrary topological space (e.g with the fixpoint property), then $X$ is a retract of  $X\times\mathbb R$, which does not have the fixpoint property (as can be seen from the fixpoint-free map $(x,t)\mapsto (x,t+1)$).
b) see a)
c) Let $Z=X\cup Y$ where $X\cap Y=\{p\}$ and $f\colon Z\to Z$ continuous, and $X,Y$ have the fixedpoint property. Then wlog $f(p)\in X$. Then $$x\mapsto\begin{cases}f(x)&\text{if $f(x)\in X$}\\p&\text{if $f(x)\notin X$}\end{cases}$$ is a continuous map $X\to X$, hence has a fixedpoint $a$. So either $f(a)=a\in X$ or $f(a)\notin X$ and $a=p$. But the latter is impossible as we assumed $f(p)\in X$. Hence $a$ is a fixedpoint of $f$.
d) Let $X$ be the ball $B=\{\,x\in\mathbb R^n\mid |x|\le 1\,\}$ with the point $s\in \partial B$ removed. Perform a shear mapping $\phi$ that leaves $s$ (and the tangential plane through $s$) fixed. Then for each $x\in X$ let the line through $s$ and $\phi(x)$ intersect $\partial B$ in $a(x)$ and $\phi(\partial B)$ in $b(x)$ (apart from $s$, of course). Then the map $$x\mapsto\frac{|a(x)-s|}{|b(x)-s|}(\phi(x)-s) + s$$ is a fixedpoint-free homeomorphism $X\to X$.
A: a) Clearly not, since $[0,1]$ is a retract of $\mathbb R$.
b) See the answer to a)
c) English is not my first language so I don't know what a "one point union" is
d) I don't think the unit ball with a pole removed has the fixed point property, but I am not certain. I think it should be possible to rotate the ball around an axis entering through the missing point, then just skewing it a little to remove all fixed points.
