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In the topology course I attend we work with the following, rather unusual definition of homology groups:

For topological spaces $X,Y$ define the presheaf $\mathcal{F}_Y$ on $X$ as follows: Let $\mathcal{F}_Y(U)$ be the free abelian group generated by the set of continuous functions from $U$ to $Y$ and the restriction is just given by restricting the basis elements.

Let $\Phi\mathcal{F}_Y$ be the sheafification of $\mathcal{F}_Y$. We call the elements of $\Phi\mathcal{F}_Y$ "mapping cycles from $X$ to $Y$". Mapping cycles can be composed in the natural way and every map $f\colon X\to Y$ induces a mapping cycle $[f]$ from $X$ to $Y$. Two mapping cycles $f,g$ from $X$ to $Y$ are said to be homotopic, if there is a mapping cycle $H$ from $X\times[0,1]$ to $Y$ such that $H_0=f,H_1=g$. Here, $H_0=H\circ [i_0]$, where $i_0\colon X\to X\times\{0\}\subset X\times[0,1]$ is the inclusion and for $H_1$ analogously. The set of homotopy classes of mapping cycles from $X$ to $Y$ carries a natural group structure. We denote it by $[[X,Y]]$.

Now, we can finally define the homology group of $X$ as $H_n(X)=[[S^n,X]]/[[\ast,X]]$, where the embedding of $[[\ast,X]]$ in $[[S^n,X]]$ is induced by the inclusion $\ast\to S^n$.

My question: We have to show that $H_n(X)\times H_m(X)\times H_{m+n}(X)\cong [[S^m\times S^n,X]]/[[\ast,X]]$ for $m,n>0$.

I assume one can prove this by writing down an isomorphism explicitly and checking all the details by hand. However, that doesn't sound like fun - there are several equivalence relations involved in the definition that are quite technical. On the other hand, sheafification and the free abelian group have nice universal properties. So, my real question is:

Is there an abstract, not too technical way to see the above isomorphism?

Thanks in advance.

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