Whitehead product and a homotopy group of a wedge sum Note : this is a crosspost from the Mathematics StackExchange, as suggested by this meta post.
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 ?
 A: Here are some details which are related to Tyler's comment.
I recommend looking at the paper "Induced Fibrations and Cofibrations" by Tudor Ganea (1967).  For connected based spaces $X$ and $Y$, there is a fibration up to homotopy
$$
\Sigma (\Omega X) \wedge (\Omega Y) \to X\vee Y \to X\times Y
$$
where the first map in the display is a kind of generalized Whitehead product (see below).
After looping once, the sequence splits, so
$$
\Omega (X\vee Y)  \simeq \Omega X \times \Omega Y \times \Omega \Sigma ((\Omega X) \wedge (\Omega Y))\, .
$$
Your isomorphism will follow by applying $\pi_{n+k}$ to this splitting--we only need to identify the term on the right.
To this end,
note that if $X$ is $n$-connected and $Y$ is $k$-connected ($n,k\ge 1$), then $\Omega \Sigma ((\Omega X)\wedge (\Omega Y))$ is $(n+k-1)$-connected (here I am using the Hurewicz theorem).  Moreover, the map
$$
(\Omega X)\wedge (\Omega Y)\to \Omega \Sigma (\Omega X)\wedge (\Omega Y)
$$
is $(2n+2k-1)$-connected.  In particular, it will induce an isomorphism
on $\pi_{n+k}$.
As $(\Omega X)\wedge (\Omega Y)$ is $(n+k-1)$-connected, the Hurewicz theorem says that
$$
\pi_{n+k} ((\Omega X)\wedge (\Omega Y)) \cong H_{n-k} ((\Omega X)\wedge (\Omega Y))
$$
and the Künneth formula provides an isomorphism
$$
H_{n+k} ((\Omega X)\wedge (\Omega Y)) \cong H_n((\Omega X) \otimes H_k(\Omega Y)\, .
$$
Another application of the Hurewicz theorem shows that
$$
H_n(\Omega X) \otimes H_k(\Omega Y) \cong \pi_{n+1}(X) \otimes \pi_{k+1}(Y)\, .
$$
Putting this all together, we  obtain an isomorphism
$$
\pi_{n+k+1} (X\vee Y)   \cong \pi_{n+k+1} (X)  \oplus\pi_{n+k+1} (Y) 
\oplus \, \,  \pi_{n+1}(X) \otimes \pi_{k+1}(Y)\, .
$$\

It remains describe the generalized Whitehead product
$\Sigma (\Omega X) \wedge (\Omega Y) \to X\vee Y $.
Taking the adjoint, we seek a map
$$
 (\Omega X) \wedge (\Omega Y) \to \Omega(X\vee Y)\, .
$$
Now, there are evident inclusions $ \Omega X \to \Omega(X\vee Y)$
and $\Omega Y  \to \Omega(X\vee Y)$. Very roughly, the idea is to map
a pair of loops $(\gamma,\omega) \in (\Omega X) \wedge (\Omega Y) $ to
the commutator
$$
[\gamma,\omega] \in \Omega(X\vee Y)
$$
where care is required to make sense of the commutator. I will refrain from writing down the formula here. I believe that the details may be found
in Ganea's paper.
