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I'm reading Rotman's Introduction to Algebraic Topology and he states that the set of morphisms between two chain complexes in the category $\text{Comp}$ (defined in the box below) is in fact an abelian group.

$(S_*, \partial) \in \text{Obj Comp}$, where $(S_*, \partial)$ means chain complex.

Hom$(S_*', S_*) = \{ \text{Set of all chain maps between $S_*'$ and $S_*$}\}$

Composition is defined as: $\{f_n\} \circ \{g_n \} = \{f_n \circ g_n\}$, for chain maps $f=\{f_n\}$, and $g=\{g_n\}$.

He states that if $f=\{f_n\}$ and $g=\{g_n\} \in $ Hom $(S_*', S_*)$, then , $f + g \in$ Hom$(S_*', S_*)$. I'm trying to prove to myself that this is a group, but what does this operation represent? What is $\{f_n\} + \{g_n\}$?

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$\{f_n\} + \{g_n\}$ is the family of morphism of abelian groups $S'_n \to S_n, x \mapsto f_n(x) + g_n(x)$.

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  • $\begingroup$ What does $f_n(x) + g_n(x)$ mean? $\endgroup$ – Oliver G Jun 14 '17 at 14:22
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    $\begingroup$ By definition $S_n$ is an abelian group (or a $R$-module depending of your definition) right ? $\endgroup$ – user171326 Jun 14 '17 at 14:23
  • $\begingroup$ Yes, $S_n$ is a free abelian group. So if $f,g: X \rightarrow Y$ and $f_n ,g_n: S_n(X) \rightarrow S_n(Y)$, then $f_n(x \in S_n(X)) + g_n(x \in S_n(X)) = \sigma_{f_n(x)} + \sigma_{g_n(x)}$ where $\sigma \in S_n(Y)$? $\endgroup$ – Oliver G Jun 14 '17 at 14:27
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    $\begingroup$ I don't really understand what do you mean by $\sigma$. The situation is simple : you have morphism $f_n, g_n : S_n \to S_n'$. You define a morphism $(f_n + g_n)(x)$ as $(f_n + g_n)(x) = f_n(x) + g_n(x)$. It is by immediate definition a morphism of groups (and also a chain complex morphism). $\endgroup$ – user171326 Jun 14 '17 at 14:31
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    $\begingroup$ Yes exactly, and this how you can define the sum of two morphisms of chain complex. $\endgroup$ – user171326 Jun 14 '17 at 14:38
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$$\{ f_n \} + \{ g_n \} = \{ f_n + g_n \} $$

That is, the $n$-th component of the sum of chain morphisms is the sum of the $n$-th components of the two terms.


If you've never seen the a sum of morphisms, then I will presume you are working in the special case of chain complexes of abelian groups.

Addition is a standard operation on abelian group homomorphisms, and is defined pointwise: if $u,v : G \to H$ are abelian group homomorphisms, then $u+v : G \to H$ is a group homomorphism as well, with values given by

$$ (u + v)(g) = u(g) + v(g) $$

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  • $\begingroup$ Why did you answer since another answer was already accepted ? (Don't take it as aggressiveness, I am just curious). $\endgroup$ – user171326 Jun 15 '17 at 19:56

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