I'd like to prove that if $n$ is even $\exists [f] \in \pi_{2n-1}(\mathbb{S}^n)$ such that $H(f) = \pm 2$. Here $H(f)$ denotes the Hopf invariant of $f$.
Sketch of the proof: Think $\mathbb{S}^n \vee \mathbb{S}^n$ as the $(2n-1)$-skeleton of $\mathbb{S}^n \times \mathbb{S}^n$.
Let $g$ be attaching map of a $(2n)-$cell, i.e $g: \mathbb{S}^{2n-1} \longmapsto \mathbb{S}^n \vee \mathbb{S}^n$. Let $h : \mathbb{S}^n \vee \mathbb{S}^n \longmapsto \mathbb{S}^n$ defined by $h(x,\star) := (\star, x) := x$ and $f:= h \circ g$.
We have that $C_g = \mathbb{S}^n \vee \mathbb{S}^n \cup (2n)-\text{cell} = \mathbb{S}^n \times \mathbb{S}^n$
First question: Shouldn't hold that $C_g$ is a quotient? I can't see why is equal to $\mathbb{S}^n \vee \mathbb{S}^n \cup (2n)-\text{cell}$.
Now, there's the following commutative diagram:
$$\begin{array}{ccccccccc} H^n(\mathbb{S}^n \vee \mathbb{S}^n) & \overset{j_{g}^*}{\leftarrow} & H^{n}(C_g) & \overset{\partial_g}{\leftarrow} & H^{n-1}(\mathbb{S}^{2n-1}) = 0 \\\ \uparrow{h^*} & & \uparrow{\tilde{h}^*} & & \uparrow{id^*} \\\ H^n(\mathbb{S}^n) & \overset{j_{y}^*}{\leftarrow} & H^n(C_f) & \overset{\partial_f}{\leftarrow} & H^{n-1}(\mathbb{S}^{2n-1})=0 \end{array}$$
Question 2: I don't get why the diagram is commutative, since $\tilde{h}$ doesn't seem to be defined.
In particular I'm having trouble computing $h^*(y)$, it should hold that $$(j_y^{*})^{-1}(h^*(y)) = (j_y^{*})^{-1}((y,0)+(0,y)) = (y \otimes 1) + (1 \otimes y)$$
But I'm unable to verify it explicitly.
Any help or reference woud be appreciated.