# 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 ?

• I don't recall the proof exactly, but it might help to know that the homotopy fibre of $X\vee Y\to X\times Y$ is $\Omega X\ast \Omega Y$ (the join of the loop spaces). Apr 23 at 20:00
• To proceed you probably first need an excision-type result, identifying $\pi_{n+k+2}(X \times Y, X \vee Y)$ with $\pi_{n+k+2}(X \wedge Y)$. Then you can use the Hurewicz theorem + Kunneth formula to identify $\pi_{n+k+2}(X \wedge Y)$ with $\pi_{n+1}(X) \otimes \pi_{k+1}(Y)$. Apr 23 at 20:14
• Look up the Hilton-Milnor theorem (or really just Hilton's 1955 paper). Apr 24 at 21:20

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.

• I must say, I had tried the argument without much success, so thank you very much for detailing it ! Apr 27 at 8:43