# Whitehead product and a homotopy group of a wedge sum

Note : this question has been crossposted on the mathematics Overflow.

Let $$X$$ be an $$n$$-connected ($$n\geqslant1$$) CW-complex and $$Y$$ be a $$k$$-connected ($$k\geqslant1$$) CW-complex. My goal is to prove the following isomorphism : $$\pi_{n+k+1}(X\vee Y)\cong\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\oplus[\pi_{n+1}(X),\pi_{k+1}(Y)],$$ with $$[\;\cdot\;,\;\cdot\;]$$ denoting the Whitehead product (here, it is understood that we take the whitehead product of the subgroups $$\pi_{n+1}(X)<\pi_{n+1}(X\times Y)$$ and $$\pi_{k+1}(Y)<\pi_{k+1}(X\times Y)$$).

So far, I have done the following. (Do let me know if I have done any mistake !)

We can always assume, up to a homotopy equivalence, by the hypothesis on $$X$$ and $$Y$$, that their respective $$n$$ and $$k$$ skeletons are of the following form : $$\text{Sk}_nX=\{\ast\}\qquad\text{and}\qquad\text{Sk}_kY=\{\ast\}.$$ In particular, $$X$$ and $$Y$$ only have cells in dimensions $$\geqslant n+1$$ and $$\geqslant k+1$$ respectively. Therefore, the product $$X\times Y$$ has only cells starting in dimension $$n+1$$ or $$k+1$$, accordingly to which one is the smallest, and that cells in dimensions $$\leqslant n+k+1$$ come from cells of either $$X$$ or $$Y$$, but not both. Therefore, we get : $$\text{Sk}_{n+k+1}(X\times Y)\subset X\vee Y,$$ and thus the pair $$(X\times Y,X\vee Y)$$ is $$(n+k+1)$$-connected.

I then tried using a part of the exact sequence of the pair :

$$\dots\longrightarrow\pi_{n+k+2}(X\times Y,X\vee Y)\overset{\partial_\ast}{\longrightarrow}\pi_{n+k+1}(X\vee Y)\overset{\imath_\ast}{\longrightarrow}\pi_{n+k+1}(X\times Y)\overset{\text{rel}_\ast}{\longrightarrow}\pi_{n+k+1}(X\times Y,X\vee Y)\longrightarrow\dots$$

We can use the $$(n+k+1)$$-connectedness of the pair to re-write the sequence as :

$$\dots\longrightarrow\pi_{n+k+2}(X\times Y,X\vee Y)\overset{\partial_\ast}{\longrightarrow}\pi_{n+k+1}(X\vee Y)\overset{k}{\longrightarrow}\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\overset{\text{rel}_\ast}{\longrightarrow}0,$$

with $$k$$ being given by the composite of $$\imath_\ast$$ and of the isomorphism $$\pi_\bullet(X\times Y)\cong\pi_\bullet(X)\oplus\pi_\bullet(Y)$$.

Now, the sequence splits at $$\pi_{n+k+1}(X\vee Y)$$, since we have $$p\circ\imath=\text{id}$$ and $$q\circ\imath=\text{id}$$ in : $$X\vee Y\overset{\imath}{\longrightarrow}X\times Y\overset{p}{\longrightarrow}X\subset X\vee Y\qquad\text{and}\qquad X\vee Y\overset{\imath}{\longrightarrow}X\times Y\overset{q}{\longrightarrow}Y\subset X\vee Y,$$

by functoriality and by using that $$\pi_\bullet$$ sends products to products. We shall denote as $$p_\ast\oplus q_\ast:\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\to\pi_{n+k+1}(X\vee Y)$$ the splitting retraction. Therefore, by an algebraic lemma (not exactly the Splitting lemma, but something rather similar), we obtain : $$\pi_{n+k+1}(X\vee Y)\cong\text{Im}(p_\ast\oplus q_\ast)\oplus\ker(k).$$

Now, I recognized that $$\text{Im}(p_\ast\oplus q_\ast)\cong\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)$$ by construction, so I am left with computing $$\ker(k)$$. And here, I am completely stuck... How to recognize the Whitehead product as the kernel I am missing ?

• What you've done so far looks good. By exactness of the (first) sequence $\ker(k)\cong \pi_{n+k+2}(X\times Y,X\vee Y)$. Since $\pi_{n+k+1}(X\times Y,X\vee Y)=0$, the Hurewicz map gives an isomorphism $H_{n+k+2}(X\times Y,X\vee Y)\cong \pi_{n+k+2}(X\times Y,X\vee Y)$, and from here the relative Künneth formula gives $H_{n+k+2}(X\times Y,X\vee Y)\cong H_{n+1}(X,\ast)\otimes H_{k+1}(Y,\ast)$. Of course apply Hurewicz again to get $H_{n+1}(X,\ast)\otimes H_{k+1}(Y,\ast)\cong \pi_{n+1}(X)\otimes \pi_{k+1}(Y)$. I don't see how to easily find the Whitehead product in all this, though. – Tyrone Nov 18 '20 at 19:29
• @Tyrone Hey thank you already for your comment ! I had thought about doing this with Künneth's formula, but yeah, I also didn't manage to get the Whitehead product out of this... Or at least it's not evident that this tensor product is what we are looking for ! – Anthony Saint-Criq Nov 18 '20 at 21:56