Various possible modification of fixed point property

Question

I was thinking about of possible variation of the fixed point property, as in the title. A space $X$ satisfies the fixed point property (FPP) if every continuous function $f:X\to X$ has a fixed point, i.e. $\exists x\in X,f(x)=x$. I realized that, for example, the space $X=[0,1)$ doesn't clearly satisfies FPP, BUT every surjective continuous function from $X$ to itself has a fixed point ($\exists xf(x)=0\leq x$ and $f(0)>0$). We can call this property SFPP (surjective FPP) and, more in general, if we select a class $\mathcal{A}$ of continuous functions from $X$ to itself, let's define $\mathcal{A}$-FPP the property: "every $f\in \mathcal{A}$ has a fixed point".

So we came to the question: what implications hold between FPP, SFPP, IFPP (injective FPP), BFPP (bijective FPP) (note that it's different from SFFP $\cap$ IFPP, at least in principle) and HFPP (homeomorphism FPP)?

Clearly FPP $\Rightarrow$ SFPP/IFPP $\Rightarrow$ BFPP $\Rightarrow$ HFPP, but what other implications hold? In particular between SFPP and IFPP.

And, even more important, also why the others doesn't hold? The previous example shows SFPP $\nRightarrow$ FPP, and also a "8"-shape (or two intersecting straight line) space shows BFPP $\nRightarrow$ FPP. Clearly an example that shows HFPP $\nRightarrow$ BFPP (if it's exists at all) cannot be compact.

EDIT. I would also like to know if BFPP=SFFP $\cap$ IFPP (probably not, but I can't think to a counterexample).

Answer 1

Partial answer covering some cases, I'll think about the rest and update it if I come up with more examples.

$SFPP\not\Rightarrow IFPP$ by your $[0,1)$ example.

$HFPP\not\Rightarrow BFPP$ by the following example: Consider the subspace of $\Bbb R^2$ composed of the $x$ axis and, for every integer $z\leq 0$ attach to $(z,0)$ a vertical half open segment, while for $z>0$ attach to $(z,0)$ a circle. This space has a fixed point free continuous bijection, namely shift every vertical segment and every circle to the right by $1$, and wrap the vertical segment over $(0,0)$ around the circle over $(1,0)$. However homeomorphisms of this space must preserve the basepoints of segments and circles.