Prove that $(g ◦ f)_∗ = g_∗ ◦ f_∗ : H_∗(C_∗) → H_∗(E_∗)$. Also show that $(Id_{C_∗})_∗ = Id_{H∗(C∗)}: H_∗(C_∗) → H_∗(C_∗)$.

Question

Let $C_∗, D_∗$ and $E_∗$ be the complexes of chains and $f_∗ : C_∗ → D_∗, g_∗ : D_∗ → E_∗$ be two homomorphisms of chain complexes. Prove that $(g ◦ f)_∗ = g_∗ ◦ f_∗ : H_∗(C_∗) → H_∗(E_∗)$. Also show that $(Id_{C_∗})_∗ = Id_{H∗(C∗)}: H_∗(C_∗) → H_∗(C_∗)$.

If $\overline{w}\in H_n(C_*)=Z_n(C_*)/B_n(C_*)$ then $f_n(\overline{w})=\overline{f_n(w)}$ and likewise with $g_n$. I have tried to do the following but I do not know if it is ok:

$(g\circ f)_n(\overline{w})=\overline{(g\circ f)_n(w)}=\overline{g_n(f_n(w))}=g_n(f_n(\overline{w}))$

Could anyone help me, please? Thank you very much

Answer 1

you only need to check the definition of the induced morphism between homology is well defined.

Show that $f_*:H_n(C_*)\rightarrow H_n(D_*)$ which sent $\overline {w}$ to $\overline{f_n(w)}$ is well defined.

You need to check if $\overline{w}=\overline{x}$,then$\overline{f_n(w)}=\overline{f_n(x)}$.

If you do this, what you have asked is clear.