# Why are the maps we get from the cup product linear?

As we introduced a cohomology group of a complex $$K$$ (for example over $$\mathbb{Z}$$) as the set $$Hom(C,\mathbb{Z})$$, we also talked about the cup product to give the cohomology group a ring structure.

In particular, the cup product is defined as follows: We define the cup product on the basis elements of the simplex as

$$\cup : C^n(K) \times C^m(K) \to C^{n+m}(K)$$

$$(\varphi \cup \psi)(\sigma_{n+m}) := \varphi(\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n})$$

and then take the linear map corresponding to the image of the basis elements.

I was wondering about the following observation: Since $$\varphi \cup \psi, \varphi,\psi$$ are linear maps, we get for a basis element $$\sigma$$

$$2(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)(\sigma)+(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)( \sigma + \sigma)=\varphi(\sigma_{\leq n}+\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n} + \sigma_{\geq n})=2\varphi(\sigma_{\leq n}) \cdot 2\psi(\sigma_{\geq n})=4(\varphi \cup \psi)(\sigma)$$

So the map cannot be linear anymore. Where is my mistake in this observation?

• I recommend you have a more precise title. Unfortunately I can’t confidently suggest one as I don’t know the subject, maybe: “why are the maps linear in the cup product (cohomology)?” Oct 15, 2022 at 10:17
• Thanks for your comments! Oct 15, 2022 at 10:49
• "we introduced a cohomology group of a complex $K$ as the set $Hom(C,\mathbb{Z})$". What is the relation betweeen $K$ and $C$? Is $K$ a chain complex or something else? And the cohomology groups are certainly not the groups $Hom(C,\mathbb Z)$, but the cohomology groups of the cochain complex $Hom(C, \mathbb Z)$. Oct 15, 2022 at 15:40
• Sorry, this was a typo. I meant $Hom(K,\mathbb{Z})$ Oct 15, 2022 at 16:00
• @mkfrnk Then you should correct it in the question. Oct 17, 2022 at 9:44

Your third step is incorrect. It is not true that $$(\phi \cup \psi)(2\sigma) = \phi(2\sigma_{\le n}) \psi(2\sigma_{\ge n})$$.
The problem is that $$2\sigma$$ is not a basis element. The formula you gave defined $$\phi \cup \psi$$ on basis elements and extended linearly, so by definition, $$(\phi \cup \psi)(2\sigma) = 2(\phi \cup \psi)(\sigma) = 2\phi(\sigma_{\le n}) \psi(\sigma_{\ge n})$$.
Your formula would only make sense if $$2\sigma$$ was a basis element with $$(2\sigma)_{\le n} = 2\sigma_{\le n}$$ and $$(2\sigma)_{\ge n} = 2\sigma_{\ge n}$$, but this is simply not true.