Naturality of Suspension I have a question about an argument used in the proof of a nameless lemma introduced below. Source: P. May's "A Concise Course in Algebraic Topology"; page 98). Here the excerpt:

The author claims that it  suffice to show that $\Sigma \partial_{n-1} \circ \partial $ null homotopic by naturality of $\Sigma$ ? 
In what way does it make sense? $\Sigma$ is suspension (functor) that maps top space $X$ to $\Sigma X := X \times S^1 /(\{*\} \times S^1 \cup X \times \{1\})$ (see page 57). 
I don't see how one can give $\Sigma$ meaning as a natural transformation after passing to homology; therefore applying $H_n(-)$ to the diagram.
 A: May uses the same symbol $\Sigma$ to denote the suspension functor (defined for based spaces) and the map  $\Sigma : \tilde{H}'_n(X) \to \tilde{H}'_{n+1}(\Sigma X)$. For the sake of precision let us denote the latter map by $\tilde{\Sigma}$. As Steve D remarked, $\tilde{\Sigma}$ is a natural transformation from the functor $\tilde{H}'_n$ to the functor $\tilde{H}'_{n+1} \circ \Sigma$ which are both living on the category of based $(n−1)$-connected spaces and based continuous maps between such spaces. This was not explicitly proved by May, but is fairly easy to verify.
In fact $\tilde{\Sigma} : \tilde{H}'_{n-1}(X^{n-1}/X^{n-2}) \to \tilde{H}'_n(\Sigma(X^{n-1}/X^{n-2}))$ is an isomorphism by a previous lemma which justifies to use $\tilde{\Sigma}^{-1}$ in the definition of $d_n$.
Instead of $d_{n-1} \circ d_n = 0$ May proves the equivalent $d'_{n-1} \circ d'_n = 0$.
As Tyrone remarked, $\Sigma \pi \circ \Sigma i$ is inessential by functoriality of $\Sigma$ which implies that $\Sigma \partial_{n-1} \circ \partial_n \simeq 0$ and therefore $(\Sigma \partial_{n-1})_\ast \circ (\partial_n)_\ast = (\Sigma \partial_{n-1} \circ \partial_n)_\ast = 0$.
But now $\tilde{\Sigma} \circ (\partial_{n-1})_\ast  = (\Sigma \partial_{n-1})_\ast \circ \tilde{\Sigma}$ by naturality of $\tilde{\Sigma}$ (this is how the last sentence in your screenshot should be read). Hence
$$(\partial_{n-1})_\ast \circ \tilde{\Sigma}^{-1} =  \tilde{\Sigma}^{-1} \circ (\Sigma \partial_{n-1})_\ast$$
and
$$d'_{n-1} \circ d'_n = \tilde{\Sigma}^{-1} \circ (\partial_{n-1})_\ast \circ \tilde{\Sigma}^{-1} \circ (\partial_n)_\ast = \tilde{\Sigma}^{-1} \circ \tilde{\Sigma}^{-1} \circ (\Sigma \partial_{n-1})_\ast \circ (\partial_n)_\ast = 0 .$$
