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Let us denote $$ \cdots \xrightarrow{\enspace d^A_n \enspace} A_n \xrightarrow{\enspace d^A_{n+1} \enspace} A_{n+1} \xrightarrow{\enspace d^A_{n+2} \enspace} \cdots $$ by $A_\bullet$ , and $$ \cdots \xrightarrow{\enspace d^B_n \enspace} B_n \xrightarrow{\enspace d^B_{n+1} \enspace} B_{n+1} \xrightarrow{\enspace d^B_{n+2} \enspace} \cdots $$ by $B_\bullet$.

There is a definition in my notes that says the following:

Let $A_\bullet$ and $B_\bullet$ be chain complexes. Then the map or morphism $f_\bullet \colon A_\bullet \rightarrow B_\bullet$ is a sequence $f_i \colon A_i \rightarrow B_i$ so that we have the "ladder" $$ \require{AMScd} \begin{CD} \cdots @> d_{n-1}^A >> A_{n-1} @> d_n^A > > A_n @> d_{n+1}^A >> A_{n+1} @> d_{n+2}^A >> \cdots \\ @. @VV f_{n-1} V @VV f_n V @VV f_{n+1} V @. \\ \cdots @> d_{n-1}^B >> B_{n-1} @> d_n^B > > B_n @> d_{n+1}^B >> B_{n+1} @> d_{n+2}^B >> \cdots \end{CD} $$ where all squares are commutative, i.e., for each $n$ we have $d_n f_n = f_{n-1} d_n$.

The exercise is as follows:

Show that if $f_\bullet \colon A_\bullet \rightarrow B_\bullet$ a monomorphism of complexes then $f_\bullet$ sends an $n$-cycle (resp. an $n$-boundary) of $A_\bullet$ into an $n$-cycle (resp. an $n$-boundary) of $B_\bullet$.

Suppose that $\alpha \in A_n$ is an $n$-cycle. Then $d^A_n (\alpha) = 0$ implies that $f_{n-1} \circ d^A_n (\alpha) = 0$. But since the squares commute, we have $f_{n-1} \circ d^A_n (\alpha) = d^B_n \circ f_n(\alpha)$. So $d^B_n(f_n(\alpha)) = 0$. So $f_n(\alpha)$ is also an $n$-cycle, since $d^B_n$ sends it to zero.

Do you think my answer is correct? I'm just wondering, because my answer is really short...

Thanks in advance

Jendrik Stelzner
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1 Answers1

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You need to be a little more careful with your indices. If $\alpha\in A_n$ is an $n$-cycle, then $d_{n+1}^A(\alpha)=0$ and so $$(f_{n+1}\circ d_{n+1}^A)(\alpha)=f_{n+1}(d_{n+1}^A(\alpha))=f_{n+1}(0)=0.$$ By commutativity we then get $$d_{n+1}^B(f_n(\alpha))=(d_{n+1}^B \circ f_n)(\alpha)=(f_{n+1}\circ d_{n+1}^A)(\alpha)=0$$ and hence $f_n(\alpha)$ is an $n$-cycle as it lies in the kernel of $d_{n+1}^B$.

This method of proof is known as diagram chasing. I'm sure you can see why. Were you ok with proving the second part of the question (the invariance of boundaries under monomorphisms)?

Dan Rust
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  • Thanks a lot. Actually, the arrows in my notes are left arrows, but I accidentally wrote them as right arrows here...which is why I mixed up the indices. Also, thank you for pointing out the second part, because I actually missed it. So if we think about the right arrows as left arrows...this is what I said: Let $\beta' \in A_n$ and $\beta \in A_{n+1}$. Assume that $d^A_{n+1} (\beta) = \beta'$. If I can show that $d^B_{n+1}(f_{n+1}(\beta)) = f_n(\beta')$, I will be done. But $f_n(\beta') = f_n(d^A_{n+1}(\beta))=d^B_{n+1}(f_{n+1}(\beta))$ by commutativity. Is that right? –  Oct 01 '13 at 00:51
  • That also looks fine (again the indices seem to be off so just make sure you know where each elemenet and images of elements are coming from and going to.) I'm a little worried that the monomorphism condition of $f_{\bullet}$ wasn't used. Was there more to the question? – Dan Rust Oct 01 '13 at 01:10
  • Thank you. There was a question before this that was related to the monomorphism condition. So maybe that's why it says that it is a monomorphism. –  Oct 01 '13 at 01:16