A question on fixed point theory I asked this  question in MSE, but I did not received any answer, so I repeat it here:
https://math.stackexchange.com/questions/858238/a-question-on-fixed-point-property
Assume that $0<k<n-1$, Note that $\mathbb{C}P^{k}$ can be considered as a  closed subset of $\mathbb{C}P^{n}$, in a natural way. We collapse $\mathbb{C}P^{k}$ to  a point. The resulting space is  denoted by $\mathbb{C}P^{n}/\mathbb{C}P^{k}$ 
My fixed point question:

Does $\mathbb{C}P^{n}/\mathbb{C}P^{k}$ satisfies fixed point property?(At least when $n$ is even)

This question is  motivated by:
https://math.stackexchange.com/questions/845057/show-mathbbcp2-cp1-is-not-a-retract-of-mathbbcp4-cp1#comment1754879_845057
 A: Here is a partial affirmative answer using mod 2 Steenrod operations; the simplest case of this (for $n$ and $k$ even) is just a correction of the slightly incorrect answer originally posted by Włodzimierz Holsztyński.  The result is that if $k+1$ and $n+1$ are both odd multiples of $2^d$ for some integer $d\geq 0$, then $\mathbb{C}P^n/\mathbb{C}P^k$ has the fixed point property.  In particular, for $d=0$ we get the fixed point property whenever $n$ and $k$ are both even.   All cohomology in this answer will have coefficients in $\mathbb{F}_2$.
Let's start by describing the action of the Steenrod squares on the cohomology $H^*(\mathbb{C}P^n)=\mathbb{F}_2[x]/(x^{n+1})$.  The following formulas are easy to prove by induction using the Cartan formula (induct on $d$ and for fixed $d$ induct on $m$):
$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=x^{2^d(m+1)} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=0 \text{ if $m$ is even}$$
From these, we deduce the following for all $0\leq \ell<2^d$:
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=x^{2^d(m+1)+\ell} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=0 \text{ if $m$ is even}$$
The quotient map $\mathbb{C}P^n\to\mathbb{C}P^n/\mathbb{C}P^k$ identifies $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$ with the subring of $H^*(\mathbb{C}P^n)$ which has as a basis $\{1,x^{k+1},x^{k+2},\dots, x^n\}$, and so the same relations hold in $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$.
Suppose now that $f:\mathbb{C}P^n/\mathbb{C}P^k\to\mathbb{C}P^n/\mathbb{C}P^k$ is any map.  For $k<i\leq n$, let $a_i\in \mathbb{F}_2$ be such that $f^*(x^i)=a_ix^i$.  By the Lefschetz fixed point theorem, $f$ must have a fixed point if $1+\sum_{k+1}^n a_i\neq 0$ (the $1$ coming from $H^0$), or equivalently if $\sum a_i=0$.
Since $f^*$ must commute with Steenrod operations, we must have $a_{2^dm+\ell}=a_{2^d(m+1)+\ell}$ for $m$ odd and $0\leq \ell<2^d$, as long as $k<2^dm+\ell<2^d(m+1)+\ell\leq n$.   Together, these relations imply that if $m$ is odd and $k<2^dm<2^d(m+2)-1\leq n$, then all the $a_i$ for $2^dm\leq i \leq 2^d(m+2)-1$ are equal to each other (everything below $2^d(m+1)$ can be related to $2^d(m+1)$ using the smaller Steenrod squares, and everything above $2^d(m+1)$ can be related to something below it using $Sq^{2^{d+1}}$).  That is, the $a_i$ are constant in blocks of length $2^{d+1}$ starting from an odd multiple of $2^d$.
Now suppose that $k+1$ and $n+1$ are both odd multiples of $2^d$.  The numbers from $k+1$ to $n$ can be broken into blocks of length $2^{d+1}$, each starting with an odd multiple of $2^d$.  All of the $a_i$ in each block are equal to each other, and hence their sum is zero since there are an even number of them.  Thus the sum of all of the $a_i$ is zero, and so $f$ must have a fixed point.
Let me conclude with a couple remarks on this result.  First, as Włodzimierz observed, this argument works equally well for projective spaces over $\mathbb{R}$ or $\mathbb{H}$ (for $\mathbb{R}$, replace $Sq^{2^{d+1}}$ with $Sq^{2^d}$ and for $\mathbb{H}$ replace it with $Sq^{2^{d+2}}$).  Second, the condition obtained here is sufficient but not necessary for $\mathbb{C}P^n/\mathbb{C}P^k$ to have the fixed point property.  Indeed, in the comments I sketched an argument using cup products and integer coefficients rather than Steenrod squares and mod 2 coefficients which shows that the fixed point property holds when $n\gg k$ as long as either $n$ is even or $k$ is odd (note that in fact, using only mod 2 coefficients there is no hope of proving the fixed point property in cases where $n$ and $k$ have different parity).
A: I don't think $\mathbb{C}P^n/\mathbb{C}P^k$ (or $\mathbb{R}P^n/\mathbb{R}P^k$) ever has the fixed point property in the range you describe.  I haven't thought about this very long so could be wrong, but on first look I think you can construct an endomorphism with no fixed points in the following way:
Consider $\mathbb{C}P^n$ as the space of 1-dimensional complex subspaces of $\mathbb{C}^{n+1}$, and $\mathbb{C}P^k$ as 1-dim subspaces in some $k+1$ dimensional subspace.  Now every 1-dim subspace in $\mathbb{C}^{n+1}$ has $n$ naturally associated other subspaces, namely those orthogonal to it.  So to construct a continuous endomorphism, we could try to find a continuous choice of orthogonal subspace, and in order for it to pass to an endomorphism of the quotient, we just need to make sure that all the subspaces contained in $\mathbb{C}P^{k+1}$ are sent to a common subspace.  Since $k<n$, there is at least one choice of 1-dim subspace which is orthogonal to all of $\mathbb{C}^{k+1}$, so pick one and call it $V$.  Now given any 1-dim subspace $U$ which is not in $\mathbb{C}^{k+1}$, $U$ has an orthogonal projection $\bar{U}$ onto $\mathbb{C}^{k+1}$, and so we can apply the unique unitary operator which fixes the orthogonal complement $W$ of $U+\bar{U}$ ($W$ is codimension 2) and sends $\bar{U}$ to $U$.  Under this map, $V$ is sent to a subspace orthogonal to $U$.  So we have assigned to every 1-dim subspace of $\mathbb{C}^{n+1}$ an orthogonal one, in a manner which is clearly continuous on passage to $\mathbb{C}P^n$, and so gives a map which by construction passes to the quotient $\mathbb{C}P^n/\mathbb{C}P^k$ and has no fixed points.
