2

I want to show that there is no retraction from $S^1\times \bar{B^2}$ to $A$. Where A is the following:

enter image description here

Let us assume that $A$ is a deformation retract of $X$. THen by definition there exists a retraction $$r:X\rightarrow A;\,\,\,\,r(a)=a\,\,\,\forall a\in A$$In addition $r\circ i=id_A$ where $i:A\rightarrow X$ is the inclusion. Thus we can consider the morphisms $$r_*:\Pi_1(X)\rightarrow \Pi_1(A)$$ $$i_*:\Pi_1(A)\rightarrow \Pi_1(X)$$ where $r_*$ is surjective and $i_*$ is incective. Now consider also $$id_{\Pi_1(A)}=(r\circ i)_*:\Pi_1(A)\rightarrow \Pi_1(A)$$ and by a corrollary we know that $r_*\circ i_*=id_{\Pi_1(A)}$

Now I only need to find a contradiction.

Therefore I wanted to ask if someone could help me?

Thanks for your help.

user1294729
  • 2,008

1 Answers1

2

Let me fix some notation. We write $X=\Bbb S^1 \times \Bbb D^2$, where $\Bbb D^2$ denotes the closed unit disk. We have embeddings $j:\Bbb S^1 \cong \Bbb S^1\times\{0\}\subseteq X$ and $i:\Bbb S^1\cong A\subseteq X$. The first of both has a retraction $q:X\rightarrow \Bbb S^1, (x,t)\mapsto (x,0)$ which turns $(j,q)$ into a deformation retract, hence homotopy equivalence.

Now assume that $i$ admits a retraction $p$. As you noted this induces a surjection $\pi_1(X) \rightarrow \pi_1(A)$ on all homotopy groups. Precomposing with the homotopy equivalence $j$ we thus have a surjection $\pi_1(\Bbb S^1) \rightarrow \pi_1(\Bbb S^1)$. This is a surjective group homomorphism $\Bbb Z \rightarrow \Bbb Z$, which necessarily is an isomorphism.

Note that the composite $qi:\Bbb S^1 \rightarrow \Bbb S^1$ defines a nullhomotopic map, hence the zero map on homotopy groups. But at the same time it is a composite of two isomorphisms of homotopy groups, hence itself an isomorphism of homotopy groups. A contradiction.

Jonas Linssen
  • 11,016
  • You really do not need the higher homotopy groups for this answer. – Moishe Kohan Jan 22 '22 at 11:03
  • That was my point, there is no need to even mention higher homotopy groups, weak homotopy equivalences and Whitehead's theorem, you simply do not need these for the proof. – Moishe Kohan Jan 22 '22 at 11:14
  • @MoisheKohan I edited the answer. Shall we remove the comments? – Jonas Linssen Jan 22 '22 at 13:32
  • sorry I feel a bit lost in the argumentation. I somehow don't see the important points. So when I assume that there is a retraction $r:S^1\times D^2\rightarrow A$ Then $r_*=\Pi(S^1\times D^2)\rightarrow \Pi(A)$ is surjective. but then I don't see how you procede – user1294729 Jan 22 '22 at 14:27
  • $X$ is homotopy equivalent to $\Bbb S^1$ via $j$, so $j_\ast$ is sn isomorphism. $A$ is homeomorphic to $\Bbb S^1$, so it has the same fundamental group. Overall we get a surjective homomorphism from $\pi_1\Bbb S^1=\Bbb Z$ to itself. This has to be an isomorphism (note that we must send 1 to $\pm 1$)… – Jonas Linssen Jan 22 '22 at 14:31
  • sorry I think I don't see what it means exactly to be homotopy equivalent. So I mean sure we know that $X,Y$ are homotopy equivalent if there is $f:X\rightarrow Y$ and $g:Y\rightarrow X$ such that $g\circ f$ is homotpopic to $id_X$ and $f\circ g$ is homotopic to $id_Y$ but I don't see how to work with it really good since finding these $f,g$ is sometimes really hard – user1294729 Jan 22 '22 at 14:35
  • sorry could you explain it again I really don't can see through – user1294729 Jan 22 '22 at 14:51
  • But it is not in the case of $X$ and $\Bbb S^1$. I actually gave you both maps $j,q$, the composite $qj=\operatorname{id}$ doesn’t need a homotopy anymore and to find a homotopy from $jq$ to $\operatorname{id}_X$ is a very useful exercise. Hint: what is the easiest path from $(x,0)$ to $(x,t)$ in $X$ you can think of? – Jonas Linssen Jan 22 '22 at 14:53
  • @PrudiiArca: Here's another way of thinking about this answer. If there is a retraction $r$, then the composition $r\circ i$ is the identity on $A$, hence the map $r_\ast \circ i_\ast :\pi_1(A)\rightarrow \pi_1(A)$ is the identity, so $i_\ast$ is injective. The map $q$ is a homotopy equivalence (with homotopy inverse $j$), so $q_\ast$ is injective. Thus, since $i_\ast$ is injective, so is $q_\ast \circ i_\ast$. However, the map $q\circ i:S^1\rightarrow S^1$ clearly has winding number $0$, so $q_\ast\circ i_\ast$ is the $0$-map, so is not injective. Thus, we've reached a contradiction. – Jason DeVito - on hiatus Jan 22 '22 at 19:41
  • so sorry I don't understand both of your ways. I have now the following way posted above maybe you can take a look – user1294729 Jan 22 '22 at 19:58