I'm reading how May's "A Concise Course in Algebraic Topology" and I have some questions about following proof: LEMMA (see pages 97-99). Here the excerpt:
We have the wedge product $X= \vee_i S^n$
- How he reduces the case to one sphere? I suppose to get this one needs an isomorphism $\tilde{H}' _n (\vee_i S^n) \cong \oplus_i \tilde{H}' _n ( S^n)$ and resp $\tilde{H}' _{n+1} (\Sigma \vee_i S^n) \cong \oplus_i \tilde{H}' _{n+1} ( \Sigma S^n)$
How to obtain it or did the author had another argument in mind?
- I guess that the case $\tilde{H}' _n ( S^n) \cong \tilde{H}' _{n+1} ( \Sigma S^n)$ is then just Freudenthal? I'm a bit curriuos why that isn't mentioned here.
Assume that $n\geq 2$ and give $S^n$ the CW structure with one $0$-cell and one $n$-cell. Then the $k$-fold product $\prod^k S^n$ naturally has the product cell structure with $k\choose r$ cells of dimension $r\cdot n$ for each $0\leq r\leq n$. In particular its $n$-skeleton has one $0$-cell and $k$ n-cells and is precisely the $k$-fold wedge $\bigvee^k S^n$.
The next cells of $\prod^k S^n$ occur in dimension $2n$, so in particular the inclusion of the $n$-skeleton
$$j:\bigvee^kS^n\hookrightarrow \prod^kS^n$$
is a $(2n-1)$-connected map. Since we are working under the assumption that $n\geq 2$, we have $2n-1\geq n+1$, and so the map on homotopy groups induced by $j$ is an isomorphism in degree $n$
$$j_*:\pi_n\left(\bigvee^kS^n\right)\xrightarrow{\cong}\pi_n\left(\prod^kS^n\right).$$
The definition of $n$-connectedness of a space, and more generally a map, appears in section 10.4 - The Cellular Approximation Theorem of May's book. In deriving the isomorphism above we haven't used any machinery more advanced than basic CW theory, and if you are uncomfortable with anything here, then I would suggest a thorough reading of that entire chapter.
It now follows directly from the definitions that
$$\tilde H'_n\left(\bigvee^kS^n\right)\stackrel{def}{=}\pi_n\left(\bigvee^kS^n\right)\xrightarrow{\cong}\pi_n\left(\prod^kS^n\right)\cong \bigoplus^k\pi_n(S^n)\stackrel{def}{=}\bigoplus^k\tilde H'_n(S^n)$$
is a free abelian group of rank $k$. The second isomorphism follows from the general fact that $\pi_r(X\times Y)\cong\pi_k(X)\oplus\pi_k(Y)$. Since the product is finitely indexed it becomes a direct sum rather than product in the category of abelian groups. Moreover this isomorphism is given by sending a homotopy class $\alpha:S^n\rightarrow \prod^kS^n$ to the collection $(pr_a\circ\alpha)_{a\leq k}$, where $pr_a$ is the projection onto the $a^{th}$ factor. The point of this is that it tells us that $\pi_n\left(\prod^kS^n\right)$ is generated by the inclusions $i_a:S^n\hookrightarrow\bigvee^kS^n\hookrightarrow \prod_kS^n$, and so $\tilde H'_n\left(\bigvee^kS^n\right)=\pi_n\left(\bigvee^kS^n\right)$ must be generated by the inclusions $i_a:S^n\hookrightarrow\bigvee^kS^n$. Here $i_a:S^n\hookrightarrow \bigvee^kS^n$ is the inclusion of the $a^{th}$ wedge summand.
Now observe there is a canonical homeomorphism
$$\Sigma\left(\bigvee^kS^n\right)\cong \bigvee^k\Sigma S^n\cong \bigvee^kS^{n+1},$$
which is such that $\Sigma i_a$ identifies exactly with the inclusion $i'_a:S^{n+1}\hookrightarrow \bigvee^kS^{n+1}$ of the $a^{th}$ wedge summand.
Now still assuming $n\geq 2$ we return to the previous isomorphism and replace $n$ with $n+1$. This gives that $\tilde H'_{n+1}\left(\bigvee^kS^{n+1}\right)$ is free abelian of rank $k$ and is generated by the inclusions
$$i'_a=\Sigma i_a$$
which are in the image of the suspension homeomorphism. Since the domain of the suspension $\tilde H'_n\left(\bigvee^kS^n\right)$ is free abelian of rank $k$, and it takes generators to generators, the suspension is an isomorphism. Hence we answer your second question.
It remains now to wrap up the case of $n=1$. One first uses the generalised Van Kampen theorem to show that the fundamental group of $\bigvee^kS^1$ is the $k$-fold free product $\pi_1(\bigvee^kS^1)\cong\ast^k\mathbb{Z}$, and is again generated by the inclusions $i_a:S^1\hookrightarrow \bigvee^kS^1$. The abelianisation of this group is the free abelian group $\bigoplus^k\mathbb{Z}$ of rank $k$, and this group is generated by the images of the generators $i_a$ in the quotient $\pi_1(\bigvee^kS^1)_{ab}$.
The point of this is that the suspension
$$\Sigma: \pi_1\left(\bigvee^kS^1\right)\rightarrow\pi_2\left(\bigvee S^2\right)\cong\bigoplus^k\mathbb{Z}$$
in this case is not an isomorphism, but rather an abelianisation. Given the definition $\tilde H'_1(\bigvee^kS^1)=\pi_1(\bigvee^kS^1)_{ab}$ we see that even in this case
$$\Sigma: \tilde H'_1\left(\bigvee^kS^1\right)\xrightarrow{\cong}\tilde H'_2\left(\bigvee S^2\right)\cong\bigoplus^k\mathbb{Z}$$
is an isomorphism.