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 ?

  • 5
    $\begingroup$ 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. $\endgroup$
    – Tyrone
    Nov 18, 2020 at 19:29
  • 1
    $\begingroup$ @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 ! $\endgroup$
    – Anthony
    Nov 18, 2020 at 21:56


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