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Assume that $X$ is a topological space. We say that $X$ satisfies the strong fixed point property if the graph of every surjective continuous self-map intersect the graph of every continuous self-map on $X$. For example the interval $I=[0,1]$ satisfy this property.

An equivalent definition: If two continuous self-maps $f,g$ on $X$ have non intersecting graphs, then neither $f$ nor $g$ is surjective.

Are there some examples of manifolds (with or without boundary) of higher dimension with this property? In particular, do the closed $2$_disc, or the even dimensional real or complex projective spaces satisfy this property?

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    $\begingroup$ I don't believe the closed 2-disc has the strong fixed point property. See i.sstatic.net/REdRa.png -- here the red region in the bottommost picture is taken to be the image of a surjective function, and the blue region is the image of another function. Suitably oriented, their graphs should not intersect. Apologies for the shoddy diagram. $\endgroup$
    – Linus Hamilton
    Commented Dec 22, 2015 at 0:52
  • $\begingroup$ @LinusHamilton thank you for the comment. may you more explain about the picture? $\endgroup$
    – Ali Taghavi
    Commented Dec 23, 2015 at 21:46

2 Answers 2

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Here is an example of two surjective continuous self-maps $f$ and $g$ on the closed unit disk whose graphs are disjoint.

Consider the surjective continuous maps $F(x,y):=x e^{2i\pi y }$ and $G(x,y):=(x-1)e^{2i\pi y }$ from the closed unit square $[0,1]^2$ to the closed unit disk $D$. Clearly we have $F(x,y)=G(x,y)$ at no point $(x,y)\in [0,1]^2$. If we pre-compose $F$ and $G$ with a homeomorphism $D\to [0,1]^2$, we get the self-maps $f$ and $g$ as claimed.

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    $\begingroup$ As a consequence, any product space $X\times D$ fails to satisfy the strong FPP, since the continuous surjective self-maps $1_X\times f$ and $1_X\times g$ have disjoint graps too. $\endgroup$
    – Pietro Majer
    Commented Dec 24, 2015 at 20:32
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    – Pietro Majer
    Commented Dec 25, 2015 at 12:06
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    $\begingroup$ thanks for your interesting answer. may be this idea can be used for counter example for projective spaces, since they are the quotient of D_{n} by an appropriate action on the boundary. $\endgroup$
    – Ali Taghavi
    Commented Dec 25, 2015 at 20:23
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    $\begingroup$ And what about three pairwise disjoint graphs of surjective self-maps on $D$ ? More generally, given $X$ and $Y$, what is the maximum number of pairwise disjoint graphs of surjective maps $X\to Y$ ? $\endgroup$
    – Pietro Majer
    Commented Dec 25, 2015 at 20:36
  • $\begingroup$ Very interesting question. $\endgroup$
    – Ali Taghavi
    Commented Dec 26, 2015 at 15:12
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Three related references are the following.

If $f:D^2\to D^2$ is such that the map $H^2(D^2,S^1)\to H^2(D^2,f^{-1}(S^1))$ induced in Cech cohomology is non-trivial, then the graph of any other map $g:D^2\to D^2$ intersects the graph of $f$. This is part of Theorem A in "W. Holsztynski, On the product and composition of universal mappings of manifolds into cubes, Proc Amer. Math. Soc, 58(1976), 311–314".

If $f,g:D^2\to D^2$ are commuting maps, do their graphs intersect? this question appears here: Two commuting mappings in the disk

If $f,g:D^2\to D^2$ are commuting homeomorphisms, do the graphs of $f,g$ and of the identity $1_{D^2}$ intersect? This question has been asked here: Do commuting homeomorphisms of the $2$-disk have a common fixed point?

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