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This is probably a stupid question, so I apologize in advance.

On p. 239 of Hatcher's book, he defines the cap product $C_k(X;R)\times C^l(X;R)\to C_{k-l}(X;R)$ for $k\geq l$, which he claims is $R$-bilinear, by the formula: $$ \sigma\frown\phi=\phi(\sigma([v_0\dots v_l])\sigma([v_l\dots v_k]). $$ I can see that this is $R$-linear on the cochains, but why does $$ (\sigma_1+\sigma_2)\frown \phi=\sigma_1\frown \phi+\sigma_2\frown \phi? $$

Is he $\textit{declaring}$ that it should be bilinear by using the last equation to extend the definition to all of $C_k(X;R)$?

Thank you.

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  • $\begingroup$ Yes, this is linear extension. In more detail, for every cochain $\psi$ we have a map from $k$-simplices $\sigma$ to $k-l$ chains, and then extend this map to a homomorphism on the free abelian group on the $k$-simplices, i.e. the $k$-chains. So this is just the definining universal property of the free abelian group. These kind of definitions are abundant. However, please take a look at the FAQ of MO; you will find out that these "elementary" questions do not fit here. $\endgroup$
    – Martin Brandenburg
    Commented Apr 18, 2012 at 20:44
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    $\begingroup$ math.stackechange.com would be a fine place to ask this, though! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 18, 2012 at 20:54

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