Let $X$ and $Y$ be topological spaces with $\pi_{i}(X) = \pi_{i}(Y) = 0$ for $i \geqslant n$
Is it true that $\pi_{i}(X\vee Y) = 0$ for $i \geqslant n$?
If no, is it true for $CW$ complexes?
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As Freakish mentions in the comments, there is a contractible space $X$ for which $X\vee X$ has an uncountable fundamental group.
Of course this particular example could not happen if $X$ were well-pointed, and in particular could not happen if $X$ were a CW complex. However, there still are many examples of n-anticonnected CW complexes $X,Y$ for which $X\vee Y$ has nontrivial homotopy groups in arbitrarily high dimensions.
Some simple examples can be constructed as follows. Take any well-pointed space $X$ and let $\nabla:X\vee X\rightarrow X$ be the folding map. A little dexterity computes its homotopy fibre, showing that there is a fibration sequence $$\Sigma\Omega X\rightarrow X\vee X\xrightarrow{\nabla}X.$$ If we now take an Eilenberg-Mac Lane space $X=K(A,n)$, then $\pi_iK(A,n)=0$ for $i>n$. However, from the long exact sequence of the fibration above we extract isomorphisms $$\pi_i(\Sigma K(A,n-1))\cong \pi_i(K(A,n)\vee K(A,n))$$ for all $i>n$. These groups will generally not be trivial.
For instance, take $X=K(\mathbb{Z},2)\simeq \mathbb{C}P^\infty$. Then $\Omega \mathbb{C}P^\infty\simeq S^1$ and we have a fibration sequence $$S^2\rightarrow \mathbb{C}P^\infty\vee \mathbb{C}P^\infty\rightarrow \mathbb{C}P^\infty.$$ This leaves $$\pi_i(\mathbb{C}P^\infty\vee\mathbb{C}P^\infty)\cong\pi_iS^2\cong\pi_iS^3$$ for all $i>2$.
Tyrone
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Thank you for answer! Also is there any examples if X and Y are finite CW complexes? – Egor Kosolapov Apr 14 '22 at 13:26
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@EgorKosolapov Do you have any examples of finite CW complexes so $\pi_i(X) =0$ for $i>n>1$? – Connor Malin Apr 14 '22 at 14:17
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@ConnorMalin n-dimensional torus. Also one can take any complete manifold with non positive sectional curvature – Egor Kosolapov Apr 14 '22 at 14:36
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1@EgorKosolapov Right, but that is $n=1$. For $n=1$, the wedge of two such spaces (called aspherical) is still aspherical. This can be proven by appealing to an explicit description of the universal cover. For $n>1$, there are actually no finite dimensional spaces which satisfy the criterion. – Connor Malin Apr 14 '22 at 15:08
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@ConnorMalin Thank you. Where can I find a proof of this fact (that there are no such finite dimensional spaces)? – Egor Kosolapov Apr 14 '22 at 15:49
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@EgorKosolapov It would be due to Serre; he proved simply connected spaces have infinitely many nontrivial homotopy groups. I am not sure where he proved this. – Connor Malin Apr 14 '22 at 18:17
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There is some relevant discussion here. – Tyrone Apr 14 '22 at 18:27
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Serre's paper is here (see Theorem 10 of $\S$24). – Tyrone Apr 14 '22 at 18:41