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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

Conjecture
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    It is a general fact of group theory that if you have a retraction $r:G \rightarrow A$ that $G$ is a semidirect product of the image by the kernel, if the image is normal this is a direct product. So what you want to show is that if $r$ is a topological retraction $r_*$ is a group retraction. – Connor Malin Feb 19 '19 at 00:08
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    This is a case where you want the algebra to do the work not the topology. – Connor Malin Feb 19 '19 at 00:10
  • Thanks! Indeed I am trying to use algebra. could you please provide a reference of a theorem that proves the fact that a retraction induces a semidirect product of the kernel by the image? – Conjecture Feb 19 '19 at 14:42
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    The kernel is clearly normal. The intersection is trivial because if $x \in ker(r) \cap im(r)$, then $r(x)=1$ and $r(x)=x$. Any element $g$ can be written as $r(g)(r(g)^{-1}g)$ where $r(g) \in im(r)$ and $r(g)^{-1}g \in ker(r)$ since $r(r(g)^{-1}g)=r^2(g^{-1})r(g)=r(g^{-1})r(g)=1$. Hence it is a semidirect product. – Connor Malin Feb 19 '19 at 19:18
  • @ConnorMalin It is not about $im(r)$ here, it is about $i_∗ (π_1 (A, a))$ – Conjecture Feb 19 '19 at 20:05
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    So you should show that those are the same thing. – Connor Malin Feb 19 '19 at 20:20

2 Answers2

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

Conjecture
  • 3,088
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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.

Soheil Haghighi
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