Applying the functor $H_*$ to the inclusion sequence $A\rightarrow B\rightarrow C$

Question

Does applying the functor $H_*$ to the sequence of inclusions $A\rightarrow B\rightarrow C$ induce a map $\phi_3: H_*(B)\rightarrow H_*(C )$, such that if $\phi_1:H_*(A)\rightarrow H_*(B)$, and $\phi_2:H_*(A)\rightarrow H_*(C )$, then $\phi_3\circ \phi_1=\phi_2$?

Answer 1

A functor $F:\mathcal{C}\to\mathcal{D}$ takes a composition $f\circ g\mapsto F(f)\circ F(g)$. Hence, the functor $H_\ast$ takes a sequence of inclusions $A\hookrightarrow B\hookrightarrow C$ to a sequence of maps $F(A)\to F(B)\to F(C)$. As Qiaochu says, "that's what it to be a functor".