The capture is from the book "R.M.Switzer, Algebraic Topology-Homology and Homotopy".
I want to know the homotopy type of the so-called "standard" homotopy equivalence $\mu:S^m\wedge S^n\longrightarrow S^{m+n}$, where $S^m:=S^{m-1}\wedge S^1$. Let $\mu\simeq d\cdot id$ for some integer $d$, then what is the value of $d$ ?
Concretely, define $\sigma^n 1$ to be the image of the unit $1\in H^0(S^0)$ under the $n$-fold suspension isomorphism $\sigma^n: H^{0}(S^0)\to H^{n}(S^n)$, where $H^{k}(X)$ is the $k$-th reduced cohomology group of $X$ and denote by $\sigma^m 1\wedge \sigma^n1$ the reduced cross product in $H^{m+n}(S^m\wedge S^n)$. Let $\mu^*(\sigma^{m+n}1)=d\cdot \sigma^m1\wedge \sigma^n1$ for some integer $d$, where $\mu^*: H^{m+n}(S^{m+n})\longrightarrow H^{m+n}(S^m\wedge S^n)$ is the induced homomorphism. What is the value of $d$?
In general for a based space $X$ there is a homeomorphism $\Sigma X\cong S^1\wedge X$. In the case that $X=S^n$ this is orientation preserving: $S^1\wedge S^n\cong S^{n+1}$ sends $\sigma^1\wedge \sigma^n$ to $+\sigma^{n+1}$. Why is this true? Well, either direct inspection, by writing $S^{n+1}=D^{n+1}\cup_{S^n}D^{n+1}$ and making the obvious identifications with the cones inside the suspension, or just by using the Freudenthal Theorem, which gives us
Now introduce the Hurewicz map $h$, and cite his theorem which says that $h:\pi_nS^n\rightarrow H_nS^n$ is an isomorphism for each $n\geq 1$. Then by naturality of $h$ we have the commutative diagram $\require{AMScd}$ \begin{CD} \pi_nS^n@>\Sigma >\cong> \pi_{n+1}S^{n+1} \\ @VhV \cong V @Vh V\cong V\\ H_nS^n @>\Sigma>\cong > H_{n+1}S^{n+1}\\ \end{CD} which tells us indeed that $\sigma^1\wedge\sigma^n$ is sent to $+\sigma^{n+1}$ under the bottom horizontal map.
Now use the associativity of the smash product $(X\wedge Y)\wedge Z\cong X\wedge (Y\wedge Z)$ (which holds for all locally compact spaces, I believe) to get
$$S^n\wedge S^m\cong (S^1\wedge\dots\wedge S^1)\wedge (S^1\wedge \dots \wedge S^1)\cong S^1\wedge(S^1\wedge(S^1\wedge(\dots \wedge(S^1\wedge S^1))\dots)$$
where after the first homoeomorphism there are $n$ copies of $S^1$ wedged together in the first bracket, $m$ copies in the second bracket. On the right-hand side you have $m+n$ copies of $S^1$, and applying the previous one factor at a time we have
$$\sigma^1\wedge(\sigma^1\wedge\dots\wedge (\sigma^1\wedge\sigma^1)\dots)=+\sigma^1\wedge(\sigma^1\wedge\dots\wedge(\sigma^1\wedge \sigma^2)\dots)=+\sigma^1\wedge(\sigma^1\wedge\dots\wedge(\sigma^1\wedge \sigma^3)\dots)$$
until we end up with
$$\sigma^1\wedge(\sigma^1\wedge\dots\wedge (\sigma^1\wedge\sigma^1)\dots)=+\sigma^{n+m}$$
On the other hand, this is just
$$\sigma^n\wedge\sigma^m=+\sigma^{m+n}$$