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I'm reading a proof to a theorem, "Suppose $a, b: H^*(-, \mathbf{Z}_2)\to H^{*+k}(-, \mathbf{Z}_2)$ are two stable (commuting with suspension isomorphism) cohomology operation of degree k. If $a(x)=b(x)$ whenever $x$ is a product of 1-dimensional classes, then $a=b$."

In the proof, it says "since $a$ and $b$ are stable it suffices to show that $a(l_n)=b(l_n)$ for $n\gg 0$, where $l_n$ is the fundamental class of $H^*(-, \mathbf{Z}_2)$." Why is it true?

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1 Answer 1

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The Yoneda lemma says that any natural transformation such as $a$ between two representable functors ($H^i(-, \Bbb Z/2\Bbb Z)$ and $H^{i+k}(-, \Bbb Z/2\Bbb Z)$ in this case) is determined uniquely by maps between the representing objects. In our case this is a map $K(\Bbb Z/2\Bbb Z, i) \to K(\Bbb Z/2\Bbb Z, i+k)$, which corresponds precisely to the element $a(l_i)$. Thus, if $a(l_i) = b(l_i)$, then $a$ and $b$ coincide on all elements in any arbitrary group $H^i(X, \Bbb Z/2\Bbb Z)$.

So far what we have done does not use the stability hypothesis at all. If we now take this into account, we see that $a,b: H^i(-, \Bbb Z/2\Bbb Z) \to H^{i+k}(-, \Bbb Z/2\Bbb Z) $ can both be rewritten as the following composition of suspension isomorphisms and shifted versions of the operations: $$ H^i(-, \Bbb Z/2\Bbb Z) \cong H^n(\Sigma^{n-i}-, \Bbb Z/2\Bbb Z) \to H^{n+k}(\Sigma^{n-i}-, \Bbb Z/2\Bbb Z) \cong H^{i+k}(-, \Bbb Z/2\Bbb Z) $$ This shows us that if $a$ and $b$ coincide on $n$-th cohomology groups for some particular $n\ge i$, they coincide on $i$-th cohomology groups. Thus, if they coincide on $n$-th cohomology groups for all sufficiently large $n$, they coincide overall.

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