4

Suppose that $X$ is a topological space, and $A$ is a retract with retraction $r: X\rightarrow A$ and $i:A\rightarrow X$ the inclusion map. Prove that if $i_*\pi(A,a)$ is normal then $$\pi(X,a)=\text{Im } i_* \times \text{Ker } r_* $$

I would like some hint to attack this problem. I don't know well how to use the $_*$ simbol. Any advice?

Thanks!

EQJ
  • 4,369

1 Answers1

2

Here's what happens: by functoriality you get that $r_*$ is onto on $\pi_1$, i.e. we have $id=(ri)_*=r_*i_* \implies r_*$ surjects. Hence you get a group isomorphism induced by $r_*$ namely: $\pi_1X/ker (r_*) \to \pi_1(A)$. In other words you could also call this an exact sequence: $0 \to ker(r_*) \to \pi_1X \to \pi_1A \to 0$. If we now could find a splitting $s:\pi_1A \to \pi_1X$, which has normal image, we would be done. Can you think of one?

Daniel Valenzuela
  • 6,305
  • 12
  • 20
  • Taking $i_*=s$. What theorem which use the normal hypotesis are you using? – EQJ Feb 13 '15 at 06:00
  • If you have a short exact sequence of non-abelian groups, you have a direct product if you have a left splitting (retraction). If you have a right splitting you need the image to be normal, then the middle term will be a product – Daniel Valenzuela Feb 13 '15 at 10:39
  • You should read the following wikipedia entry. (you always get a semidirect product out of a split short exact sequence, but (citation) "Suppose G is a semidirect product of the normal subgroup N and the subgroup H. If H is also normal in G, or equivalently, if there exists a homomorphism G → N which is the identity on N, then G is the direct product of N and H." http://en.wikipedia.org/wiki/Semidirect_product – Daniel Valenzuela Feb 13 '15 at 10:43