Let S~(ni) (i = 1,…, r) be spheres of which the topological product S~(n1) ×…× S~(nr) has a unique top dimensional cell, e~(n1+…+nr). By Y we mean the topological space obtained from S~(n1)×…× S~(nr) by removing e~(n1+…+nr). Let x_i~o, e~o and x be reference points of S~(ni), Y and X respectively. If f: Y, e~o→X, x is a continuous map and g: E~(n1+…+nr-1), E~(n1+…+nr-1)→Y, e~o represents the homotopy boundary of the element of ∏_(n1+…+nr)(S~(n1)×…× S~(nr), Y) determined by the characteristic map of e~(n1+…+nr), then fg: E~(n1+…+nr-1), E~(n1+…+nr-1)→X, x gives us an element of ∏_(n1+…+nr-1) (X, x) called the (r-1) secondary product of the elements α_i(i = 1,…, r) represented by the maps f|S~(ni): S~(ni),x_i~o→X, x. All the continuous maps of (r-1)-dimensional spheres S~(r-1) into X, carrying the reference point of S~(r-1) to x, constitute a topological space Ω_(r-1). Write E~(n1+…+nr-1)= E~(n1+…+nr-r).E~(r-1). The map fg determines a map b being the map S~(r-1)→x, such that the map satisfieswhere x ∈ E~(nx+…+nr-r) and τ ∈ E~(r-1). Evidently the map (fg) determines a homology class ξ in H_(n1+…+nr-r) (Ω_(r-1)).On the other hand we take the product of spheres, S~(n1-1)×…×S~(nr-1), and define a map Φ: S~(n1-1)×…×S~(nr-1)→Ω_(r-1) as follows: If x_i ∈ S~(ni-1), which is considered as the equator of S~(ni), then there is a unique plane, π, passing through x_i, x_i~o and perpendicular to the equator. The intersection of π and S~(n)i is a circle, namely, S_(xi)~1. The product may be considered as the image of a continuous map in natural way. A map of the sphere E~r = (E_1~1×…×E_r~1) into X is defined by if t_i and where then is continuous and We define The map Φ has a corresponding homology class (Ω_(r-1)). The purpose of this note is to prove