Assume, for $n \ge 1$, that $H_i(S^n)= Z$ if $i = 0, n$ and that $H_i(S^n) = 0$ otherwise. Prove that the equator of the $n$-sphere is not a retract, where $H_i(S^n)$ is the homology functor of $S^n$.
$(1)$ $H_n(g \circ f) = H_n(g) \circ H_n(f)$
$(2)$ $H_n(S^n)\neq 0, n \ge 0$
$(3)$ $H_n(D^{n+1})=0, n \ge 0$
$(4)$ $H_n(1_X)$ is the identity function on $H_n(X)$
This is literally the only knowledge about what a homology functor is that the book has provided me so far.
I can see that by assuming there is a retract and by assuming that $i \neq 0, n$ this leads to a commutative graph that implies $H_i(1)=H_i(r) \circ H_i(i)=0$ which clearly contradicts $(4)$.
But when $i=0,n$ I can't find a way to contradict anything based on the commutative graph:
$H_i(i) : H_i(S^{n-1}) \rightarrow \Bbb Z$
$H_i(r) : \Bbb Z \rightarrow H_i(S^{n-1})$
$H_i(1) : H_i(S^{n-1}) \rightarrow H_i(S^{n-1})$
Using solely what has been provided in the above box, anyone have any ideas?
A retract of $X$ onto $A$ is a map $r:X \to A$ such that $r \circ i=id:A \to A$, where $i:A \to X$ is inclusion.
The equatorial "circle" can be identified with $S^{n-1}$. The idea is to apply the $H_{n-1}$ functor. In this way, notice that $H_{n-1}(S^{n-1})=\mathbb Z$ and $H_{n-1}(S^n)=0$. Assume there exists a retract $r:S^{n} \to S^{n-1}$. Then $(r \circ i)_*=id_*$ by $(4)$, but by $(1)$, we also know that $(r \circ i)_*=r_* \circ i_*$. But, note that $i_*$ has to be a zero homomorphism since $H_{n-1}(S^n)$ was trivial. But then $r_*(0)=0$. How could the identity map possibly factor through a zero map? We have a contradiction.