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Exercise 1.1.4 Show that $\{\text{Hom}_R(A, C_n)\}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C_{\cdot}$. Taking $A = Z_n$, show that if $H_n(\text{Hom}_R(Z_n, C)) = 0$, then $H_n(C) = 0$. Is the converse true?

Weibel, Charles A.. An Introduction to Homological Algebra (Cambridge Studies in Advanced Mathematics) . Cambridge University Press. Kindle Edition.

Do they mean take $A = Z_n$ for some single, fixed $n$? How could they mean otherwise since the functoriality of the $\text{Hom}_R(A, \cdot)$ is only guaranteed when you hold the other argument constant. Yet they go on to talk about $H_n$ as if that safer assumption is not true.

So which is it, or is there a mathematical mistake here?

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  • $\begingroup$ Is $Z_n$ is the kernel of the map from $C_n \to C_{n-1}$ (ie the cycles)? $\endgroup$ – Tim May 31 at 19:29
  • $\begingroup$ @Tim yes that is correct $\endgroup$ – BananaCats Category Theory App May 31 at 21:36
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    $\begingroup$ I'm not sure I understand your question correctly, but I think you should fix a certain $n$ and understand $H_n(\operatorname{Hom}(Z_n,C))$ as the $n$th homology group of the complex $\{\operatorname{Hom}(Z_n,C_m)\}_m$. $\endgroup$ – Arnaud D. Jun 1 at 7:38
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As @Arnaud D mentioned, we should think of $n$ as fixed. We are applying $Hom_R(Z_n,\square)$ to the complex $C$ and then looking at the $n$th (same $n$) homology of the result.

Let the complex $C$ be as follows: $$\cdots\xrightarrow{} C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1}\xrightarrow{} \cdots$$

Apply $Hom_R(Z_n,\square)$ and assume the $n$th homology is zero. That is, for the maps $$Hom_R(Z_n,C_{n+1}) \xrightarrow{d_{n+1}^*} Hom_R(Z_n,C_n) \xrightarrow{d_n^*} Hom_R(Z_n,C_{n-1})$$ we have that $\ker d_n^*$ = Im $d_{n+1}^*$.

Now, let $y \in Z_n$. We would like to show that there exists an element $x \in C_n$ such that $d_{n+1}(x)=y$. Notice that the inclusion map $i: Z_n \to C_n$ is an element of the middle term above. Since $Z_n = \ker d_n$, we have that $d_n \circ i = 0$, that is, $i \in \ker d_{n}^*$. Since $\ker d_n^* =$ Im $d_{n+1}^*$ by assumption, there exists a map $f: Z_n \to C_{n+1}$ such that $d_{n+1}\circ f = i$. But then we are done since $y = i(y) = d_{n+1}(f(y))$.

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