*Note : this is a [crosspost](https://math.stackexchange.com/q/3907570/259363) from the Mathematics StackExchange, as suggested by [this](https://math.meta.stackexchange.com/a/23092/259363) 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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 ?



Whitehead product and a homotopy group of a wedge sum
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.


