homotopic to a constant map

Question

Let $X$ and $Y$ be topological spaces and more precisely connected finite CW complexes. Let $f\colon X \to Y$ be a continuous map such that there exist a second continuous map $F\colon X^3 \to Y$ and $$ \forall x,y\in X: F(x,x,y)=F(x,y,x)=F(y,x,x)=F(x,x,x)=f(x) $$ Does that imply that $f$ is homotopic to a constant map?

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Some partial results and motivation:

1. Consider a similar problem for groups with the usual product: If $X$ and $Y$ are groups $f\colon X \to Y$ and $F\colon X^3\to Y$ are group homomorphisms such that $F(x,x,y)=F(x,y,x)=F(y,x,x)=F(x,x,x)=f(x)$, then $$f(x)=F(x,x,x) = F(x,1,1)\cdot F(1,x,1) \cdot F(1,1,x) =(f(1))^3=1$$ and $f$ would be the zero map. This implies that $f$ induces a zero map in all homotopy groups.
2. If $X=Y$ and $f$ is the identity, the statement is true by the Whitehead theorem and the first point. This was already proven in 1977, Walter Taylor: Varieties obeying homotopy laws, Theorem 7.7
3. If $f$ is not the identity, it is not sufficent to be zero on all homotopy groups. For example the map from the torus to the sphere which contracts two circles induces a zero map in all homotopy groups, but is not homotopic to a constant map. That is my motivation: Finding a nice criteria for maps to be homotopic to a constant.
4. I have no counter example for more general $X$ and $Y$.
5. We can replace the space $Y$ by $X^3/M$ where $M$ is the equivalence relation identifying the majority. However, in general, $X^3/M$ is not contractible. It has for example nontrivial cohomology when $X$ is a circle.
6. If $X$ has dimension 1, the statement holds as this follows from the fundamental group.
7. If $X$ is a is a co-H-space in the category of pointed topological spaces, the set of maps $X\to Y$ up to homotopy becomes a group. In this case, also the statement holds. This includes the cases where $X$ is a suspension of another space. This result was brought to me by Prof. A. Thom.

Answer 1

Let $X = \Bbb{RP}^n$ for $n \geq 2$. I claim that the map $X \to X^3 / M$ is nontrivial on mod-2 cohomology.

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Here is some general material.

For $1 \leq i \leq 3$, let $p_i: X^3 \to X^2$ be the projection away from the $i$th factor, and $A_i$ the inverse image of the diagonal of $X^2$ --- eg $X_1$ is the set of points of the form $(y,x,x)$. All the spaces $A_i$ are homeomorphic to $X^2$, and their common intersections are the diagonal copy of $X$.

Let $P = A_1 \cup A_2 \cup A_3 \subset X^3$. Then there is a continuous map $g: P \to X$ such that $g(x,x,y) = g(x,y,x) = g(y,x,x) = x$, and $X^3 / M$ is the pushout of a diagram $$ X \xleftarrow{g} P \to X^3. $$ The union $P = A_1 \cup A_2 \cup A_3$ is well-behaved enough to apply the Mayer--Vietoris sequence. If we do so, we can calculate that $H^*(P)$ is isomorphic to the set of $\{(f_1, f_2, f_3)\}$ such that $f_i$ is in $H^*(A_i)$ and such that all three $f_i$ restrict to the same element in $H^*(X)$, the cohomology of the diagonal.

We can also apply the Mayer--Vietoris long exact sequence on cohomology to get an exact sequence $$ \dots \to H^*(X^3/M) \to H^*(X) \oplus H^*(X^3) \to H^*(P) \to \dots $$ Therefore, the map $H^*(X^3 / M) \to H^*(X)$ will be nontrivial if there exist a nonzero $f \in H^*(X)$ and a $g \in H^*(Y)$ such that $f$ and $g$ have the same image in $H^*(P)$. Equivalently, we need a nonzero element $f \in H^*(X)$ and a $g \in H^*(X^3)$ such that all the restrictions of $g$ to $H^*(A_i)$ are equal to restrictions of $f$ along the projection $g: A_i \to X$.

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In the specific case $X = \Bbb{RP}^{n}$, $H^*(X)$ is the polynomial ring $\Bbb F_2[x]/(x^{n+1})$. Unravelling the above identifications, we are looking for a nonzero polynomial $f(x) \in \Bbb F_2[x]/(x^{n+1})$ and an element $g(x_1,x_2,x_3) \in \Bbb F_2[x_1,x_2,x_3]/(x_i^{n+1}) \cong H^*(X^3)$ such that $(g(z,z,x_3), g(z,x_2,z), g(x_1,z,z)) = (f(z), f(z), f(z))$.

In this case, we can take $f(x) = x^2$ and $g(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$, using the fact that we are working mod $2$. This is nonzero as long as we are at least at $\Bbb{RP}^2$.

(There is no such pair of single-variable polynomials in characteristic zero.)