In Cocyclic maps and coevaluation subgroups by Lim, a map $f:X \to A$ is said to be cocyclic if there exists a map $\phi$ such that the diagram is homotopy commutative.

Is the projection $ f: S^2 \times S^3 \to S^2 $ cocyclic?

Where $S^n$ is the $n$-sphere.

I'm aware of the well known facts about the co-$H$-space structure on $S^n$ for some $n$, but I'm not sure about the product spheres. I think- if I'm not wrong- that using cup-products we can show the answer is No, but I'm not sure how to put it together.

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    $\begingroup$ I think you forgot to say which map $S^2 \times S^3 \to S^2$ you are asking about. $\endgroup$ – Omar Antolín-Camarena Mar 8 '18 at 22:02
  • $\begingroup$ @Omar Antolín-Camarena The map $f:S^2 \times S^3→S^2$ is the projection map. $\endgroup$ – Student Mar 8 '18 at 22:28

You are right about the cup product ruling this out. The cohomology ring of $X \times A = S^2 \times S^3 \times S^2$ is $\mathbb{Z}[a,b,c]/(a^2,b^2,c^2)$ where the classes $a$ and $c$ are in degree 2, coming from the two $S^2$ factors and $b$ is in degree three, from the $S^3$ factor. Then, $$H^\ast(X \vee A) = H^\ast((S^2 \times S^3) \vee S^2) = \mathbb{Z}[a,b,c]/(a^2,b^2,c^2,ac,bc),$$ with $j : X \vee A \to X \times A$ inducing the homorphism $j^\ast : a, b, c\mapsto a, b, c$.

Also, $H^\ast(X) = \mathbb{Z}[a,b]/(a^2, b^2)$ and the map $(1 \times f) \circ \Delta : X \to X \times A$ induces $a,b,c \mapsto a,b,a$ on cohomology, since you picked $f$ to be the projection. The relation $(1\times F)\circ \Delta = j \circ \phi$ for the hypothetical $\phi$, forces $\phi^\ast$ to send $a,b,c \mapsto a,b,a$, but this impossible since $bc=0$ in $H^\ast(X \vee A)$ and $\phi^\ast(bc) = \phi^\ast(b) \phi^\ast(c) = ba$ is not zero in $H^\ast(X)$.

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