Fundamental group as a product of normal subgroups

Question

Let $A$ be a retract of a non-empty topological space $X$ and let $a \in A$.

Let's denote $r : X → A$ the retraction and $i : A → X$ the inclusion.

prove that $i_∗ (π_1 (A, a))\triangleleft π_1 (X, a) \Rightarrowπ_1 (X, a)\cong i_∗ (π_1 (A, a)) × ker(r_∗)$

I proved that $ker(r_∗ ) ∩ i_∗ (π_1 (A, a)) = \{1\}$:

$ker(r_∗ ) < π_1 (X, a)$ and $i_∗ (π_1 (A, a)) < π_1 (X, a)$ so $\{1\}\in ker(r_∗ ) $ and $\{1\}\in i_∗ (π_1 (A, a))$ therefore: $\{1\} \subset ker(r_∗ ) ∩ i_∗ (π_1 (A, a))$.

Conversely, Let $\alpha \in ker(r_∗ ) ∩ i_∗ (π_1 (A, a))$. then for a $\beta \in i_∗ (π_1 (A, a))$ such that $i_∗ (\beta)=\alpha$:

$r_* \circ i_*=id_{π_1(A,a)} \Rightarrow \beta=r_* \circ i_*(\beta) = r_*(\alpha)= 1$. which proves the other inclusion.

I need to prove that

$i_∗ (π_1 (A, a))\triangleleft π_1 (X, a) \Rightarrow$

• each element of $π_1 (X, a)$ is a product of elements from $ker(r_∗ )$ and $i_∗ (π_1 (A, a))$
• $ker(r_∗ )$ and $i_∗ (π_1 (A, a)) $ commute

which will mean that the semi direct product is a direct product.

I have no clue on how to prove these two points.

Thanks for help and comments

Answer 1

$i_∗ (π_1 (A, a))\triangleleft π_1 (X, a)$ and $Ker(r_∗)\triangleleft π_1 (X, a)$ and $ker(r_∗ ) ∩ i_∗ (π_1 (A, a)) = \{1\}$ means that elements of $i_∗ (π_1 (A, a))$ and $ker(r_∗ )$ commute.

We have an exact short sequence: $ 1\to ker(r_∗ )\stackrel{\text{i}}{\to} π_1(X)\stackrel{\text{$r_*$}}{\to} π_1 (A)\to 1$ with a right section $ π_1 (A) \stackrel{\text{$i_*$}}{\to} π_1(X)$, which means, by the Splitting lemma of non abelian groups, that $ π_1 (X) \cong ker(r_*) \rtimes i_*(π_1 (A))$.

As the elements from the two sub-groups commute, this is a direct product.

Answer 2

We have this theorem in algebra:
If $N\unlhd G$ and $H\unlhd G$, then $G\cong N\times H$ if $N\cap H = \{1\}$ and $NH=G$.

Note that $\ker(r_*)$ is a normal subgroup of $\pi_1(X, a)$. And $i_*(\pi_1(A,a))$ is assumed to be normal. Also, you have shown that the intersection is trivial. So, all we need to do is to show that $\pi_1(X,a)=\ker(r_*)\ i_*(\pi_1(A,a))$.

Let $[\alpha]\in \pi_1(X,a)$. We have:

$$[\alpha] = [\alpha\ * \ \overline{r\circ\alpha}\ * \ r\circ\alpha] = [\alpha\ * \ \overline{r\circ\alpha}]\ * \ [r\circ\alpha] $$

Now $[r\circ\alpha]$ is clearly in $i_*(\pi_1(A,a))$, as $r$ sends everything into $A$. And we have

$$r_*([\alpha\ * \ \overline{r\circ\alpha}]) = [r\circ\alpha\ * \ r\circ\overline{r\circ\alpha}] = [r\circ\alpha\ * \overline{r\circ r\circ\alpha}] = [r\circ\alpha\ * \ \overline{r\circ\alpha}] = [a]$$

So, $[\alpha\ * \ \overline{r\circ\alpha}]$ is in $\ker r_*$ and we are done.