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 ?

  • $\begingroup$ 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). $\endgroup$
    – Mark Grant
    Apr 23, 2021 at 20:00
  • 1
    $\begingroup$ 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 $\pi_{n+1}(X) \otimes \pi_{k+1}(Y)$. $\endgroup$ Apr 23, 2021 at 20:14
  • 1
    $\begingroup$ Look up the Hilton-Milnor theorem (or really just Hilton's 1955 paper). $\endgroup$ Apr 24, 2021 at 21:20

1 Answer 1


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.

  • $\begingroup$ I must say, I had tried the argument without much success, so thank you very much for detailing it ! $\endgroup$
    – Anthony
    Apr 27, 2021 at 8:43

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