Deformation retract

Section 2.6 Deformation retract

Lemma 2.6.1. Homotopic maps and fundamental groups.

Let \(h,k\colon (X,x_0)\rightarrow (Y,y_0)\) be maps of pointed spaces. If there is a homotopy \(H\colon X\times I\rightarrow Y\) from \(h\) to \(k\) satisfying \(H(x_0,t)=y_0\) for all \(t\in I\text{,}\) then we have \(h_*=k_*\) as group homomorphisms from \(\pi_1(X,x_0)\) to \(\pi_1(Y,y_0)\text{.}\)

Proof.

By definition of \(h_*, k_*\text{,}\) we must show that \([h\circ f]=[k\circ f]\) for all loops \(f\in P(X; x_0,x_0)\text{:}\) equivalently, we must show \(h\circ f\simeq_p k\circ f\text{.}\) This of course is where \(H\) comes in. Define \(G\colon I\times I\rightarrow Y\) as \(G(s,t)=H(f(s),t)\text{.}\) We see that

\begin{align*} G(s,0) \amp = H(f(s),0) \amp G(s,1)\amp = H(f(s),1) \\ \amp = h(f(s)) \amp \amp =k(f(s)\\ \amp =h\circ f(s) \amp \amp =k\circ f(s)\text{,} \end{align*}

showing that \(G\) is a homotopy from \(h\) to \(k\text{.}\) It remains to show it is a path homotopy. This follows since

\begin{align*} G(0,t) \amp = H(f(0),t)=H(x_0,t)=y_0\\ G(1,t) \amp =H(f(1),t)=H(x_0,t)=y_0 \end{align*}

for all \(t\in I\text{.}\)

Theorem 2.6.2. Fundamental groups: \(S^n\) and \(\R^{n+1}-\{\boldzero\}\).

Let \(X=\R^{n+1}-\{\boldzero\}\text{,}\) let \(j\colon S^n\hookrightarrow X\) be the inclusion map, and let \(\boldx_0=(1,0,\dots, 0)\text{.}\) The function

\begin{align*} r\colon X \amp \rightarrow S^n\\ \boldx \amp \mapsto \frac{\boldx}{\norm{\boldx}} \end{align*}

is a retraction map that induces an isomorphism \(r_*\colon \pi_1(X, \boldx_0)\rightarrow \pi_1(S^n, \boldx_0)\text{.}\)

Proof.

It is clear that the given \(r\) is a retraction. Letting \(j\colon S^n\hookrightarrow X\) be the inclusion, we have \(r\circ j=\id_{S^n}\) and hence \(r_*\circ j_*=\id_{\pi_1(S^n, \boldx_0)}\text{.}\)

We are halfway to showing \(r_*\colon \pi_1(X,\boldx_0)\rightarrow \pi_1(S^n, \boldx_0)\text{.}\) It remains to show \(j_*\circ r_*=\id_{\pi_1(X,\boldx_0)}\text{,}\) and to do this we use the lemmma. Specifically we show that \(\id_X\colon X\rightarrow X\) and \(j\circ r\colon X\rightarrow X\) are homotopic via a homotopy \(H\) as described in Lemma 2.6.1. Since \(j\circ r(x)=\boldx/\norm{\boldx} \text{,}\) it is easy to see that \(H\colon X\rightarrow I\rightarrow X\)

\begin{equation*} H(x,t)=(1-t)\boldx+t\frac{\boldx}{\norm{\boldx}} \end{equation*}

is a homoto\boldx_0y from \(\id_X\) to \(j\circ r\text{.}\) Furthermore, we have

\begin{equation*} H(\boldx_0,t)=(1-t)\boldx_0+t\frac{\boldx_0}{\norm{\boldx_0}}=(1-t)\boldx_0+\boldx_0=\boldx_0\text{,} \end{equation*}

using the fact that \(\norm{\boldx_0}=1\text{.}\)

Definition 2.6.3. Deformation retract.

Let \(A\) be a subspace of \(X\text{.}\) A deformation retraction from \(X\) to \(A\) is a homotopy \(H\colon X\times I\rightarrow X\) satisfying the conditions below. (It will help to think of \(H=\{f_t\}_{t\in I}\) in the usual way.)

\(H(x,0)=x\) for all \(x\in X\) (\(f_0(x)=\id_X\));
\(H(x,1)\in A\) for all \(x\in X\) (\(f_1\colon X\rightarrow A\));
\(H(a,t)=a\) for all \(a\in A\) and \(t\in I\) (\(f_t\vert_A=\id_A\) for all \(t\in I\)).

In this case \(A\) is called a deformation retract of \(X\text{.}\)

Letting \(r\colon X\rightarrow A\) be defined as \(r(x)=H(x,1)\text{,}\) we see that \(r\) is a retraction from \(X\) to \(A\text{,}\) and \(H\) is a homotopy from \(\id_X\) to \(j\circ r\text{,}\) where \(j\colon A\hookrightarrow X\) is the inclusion map.

Remark 2.6.4. Deformation retract.

Loosely \(A\) is a deformation retract of \(X\) if there is a retraction \(r\colon X\rightarrow A\text{,}\) and this retraction is homotopic to the identity function on \(X\text{.}\)

Theorem 2.6.5. Deformation retract and fundamental groups.

Let \(A\) be a deformation retract of \(X\text{.}\) The inclusion \(j\colon A\rightarrow X\) gives rise to an isomorphism \(j_*\colon \pi_1(A,a)\rightarrow \pi_1(X,a)\) for any \(a\in A\text{.}\)

Proof.

The proof is essentially identical to the one provided for Theorem 2.6.2. As stated in the Definition 2.6.3, the map \(r\colon \)

Example 2.6.6. \(\R^3\) minus axis.

Let \(X\) be \(\R^3\) with the the \(z\)-axis removed. Show that \(\R^2-\{\boldzero\}\) is a deformation retract of \(X\text{.}\) Conclude that \(\pi_1(X, x_0)=\Z\) for any \(x_0\in X\text{.}\)

Example 2.6.7. \(\R^3\) minus circle and \(z\)-axis.

Let \(X\) be \(\R^3\) with the circle \(x^2+y^2=1\) in the \(xy\)-plane and the \(z\)-axis removed. Show that the torus \(\mathcal{S}\) obtained by revolving the circle \((x-1)^2+y^2=(1/2)^2\) is a deformation retract of \(X\text{.}\) Conclude that \(\pi_1(X,x_0)\cong \Z^2\) for any \(x_0\in X\text{.}\)

Example 2.6.8. Doubly punctured plane, figure eight, theta space.

Let \(X=\R^2-\{P\}-\{Q\}\text{,}\) where \(P=(-1/2,0), Q=(1/2,0)\text{.}\) Show that the given \(A\) is a deformation retract of \(X\text{:}\)

Figure eight.
\(A=C_1\cup C_2\text{,}\) where \(C_1\colon (x+1/2)^2+y^2=(1/2)^2\text{,}\) \(C_2\colon (x-1/2)^2+y^2=(1/2)^2\)
Theta space.
\(A=S^1\cup \{0\}\times [-1,1]\) (i.e., circle union diameter)

Prev Top Next