Show that if $\alpha_1 \sim \beta_1$ and $\alpha_2 \sim \beta_2$ , then (whenever composition makes sense) $\alpha_1 \circ \alpha_2 \sim \beta_1 \circ \beta_2$.
I have two questions.
So are these morphisms ($\alpha_1$, $\alpha_2$, $\beta_1$, and $\beta_2$) maps from $\mathcal{A}$ to $\mathcal{C}$, where $\mathcal{A}$ and $\mathcal{C}$ are chain complexes with $$\mathcal{A} := ... \leftarrow A_{n-1} \leftarrow A_n \leftarrow ...$$ and $$\mathcal{C} := ... \leftarrow C_n \leftarrow C_{n+1} \leftarrow ...$$
Or do we have $\alpha_1 : A_1 \rightarrow C_1$, $\alpha_2: A_2 \rightarrow C_2$, $\beta_1: A_1 \rightarrow C_1$, and $\beta_2: A_2 \rightarrow C_2$. I'm assuming its the former, but I'm having problems with both cases.
If its the former, then we have, for every $n$, $\alpha_{1 (n)} : A_n \rightarrow C_n$ and $\alpha_{2(n)}: A_n \rightarrow C_n$. But then how can we compose them? Because for $a \in A_n$, we have $\alpha_{1(n)} \circ \alpha_{2 (n)}(a) = \alpha_{1(n)}(\alpha_{2(n)} (a))$. But $\alpha_{2(n)}(a) \in C_n$, so its not even in the domain of $\alpha_{1(n)}$.
If $\alpha_1: A_n \rightarrow C_n$ and $\alpha_1: A_n \rightarrow C_n$, then $\alpha_1 \circ \alpha_2 (a)$ still does not make any sense.
Thanks in advance
I'm assuming that $\alpha_1,\alpha_2,\beta_1,\beta_2$ are chain complexes maps, in that case since $\alpha_1 \sim \beta_1$ e $\alpha_2 \sim \beta_2$ and so there must be $\mathcal A,\mathcal B,\mathcal C,\mathcal D$ four chain complexes such that $\alpha_1,\beta_1 \colon \mathcal A \to \mathcal B$ and $\alpha_2,\beta_2 \colon \mathcal C \to \mathcal D$.
In this way $$\alpha_1 =\langle {\alpha_1}_n \colon A_n \to B_n\rangle_{n \in \mathbb Z}$$ $$\alpha_2 =\langle {\alpha_2}_n \colon C_n \to D_n\rangle_{n \in \mathbb Z}$$ $$\beta_1 = \langle {\beta_1}_n \colon A_n \to B_n \rangle_{n \in \mathbb Z}$$ $$\beta_2 = \langle {\beta_2}_n \colon C_n \to D_n\rangle_{n \in \mathbb Z}$$
Now in order to be possible that composites $\alpha_1 \circ \alpha_2$ and $\beta_1 \circ \beta_2$ exist it must be $\mathcal A= \mathcal D$: in this way the composite maps are defined as $${\alpha_1 \circ \alpha_2}_{n} = {\alpha_1}_n \circ {\alpha_2}_n \colon C_n \to B_n$$ $${\beta_1 \circ \beta_2}_{n} = {\beta_1}_n \circ {\beta_2}_n \colon C_n \to B_n$$