Proposition

If a space $X$ retracts onto a subspace $A$, then the homomorphism \(i_*: \pi_1(A, x_0) \rightarrow \pi_1(X, x_0)\) induced by the inclusion $i: A \rightarrow X$ is injecive. If $A$ is a deformation retract of $X$, then \(i_*\) is an isomorphism.

Solution

Let $i: A \rightarrow X$ be the inclusion. Let $x_0 \in A$ be given.

Since $X$ retracts onto a subspace $A$, there must exist a continuous function $r: X \rightarrow A$ such that $r \circ i$ is the identity function on $A$.

For any $[f] \in \pi_1(A, x_0)$,

Therefore, \(r_* \circ i_*\) is the identity function on $\pi_1(A, x_0)$. Let $[f], [g] \in \pi_1(A, x_0)$ such that \(i_*([f]) = i_*([g])\). Then \([f] = (r_* \circ i_*)([f]) = r_*(i_*([f])) = r_*(i_*([g])) = (r_* \circ i_*)([g]) = [g]\). Therefore, $i_*$ is injective.

We will prove the second part. Let $A$ be a deformation retract of $X$ and $i: A \rightarrow X$ be the inclusion map. Let $x_0 \in A$ be given. Since $A$ is a retract of $X$, \(i_*: \pi_1(A, x_0) \rightarrow \pi_1(X, x_0)\) is an injective homomorphism as shown above.

We will show that \(i_*\) is surjective. Let $[f] \in \pi_1(X, x_0)$ be given. Let $r_t$ be a deformation retraction. Then we claim that $F(s, t) = r_t(f(s))$ is a homotopy between $f$ and $r_1 \circ f$.

• $F(s, 0) = r_0(f(s)) = f(s)$.
• $F(s, 1) = r_1(f(s)) = (r_1 \circ f)(s)$.
• $F(0, t) = r_t(f(0)) = r_t(x_0) = x_0$ since $x_0 \in A$.
• $F(1, t) = r_t(f(1)) = r_t(x_0) = x_0$.
• Since $r_t(s)$ is continuous when seen as a function of two variables $s, t$ and $f(s)$ is continuous, $r_t(f(s))$ is continuous.

Therefore, $F$ is indeed a homotopy. By the definition of a deformation retraction, $r_1 \circ f$ maps $I$ into $A$. Since $r_1 \circ f$ is continuous, it is a loop in $A$. Thus $[r_1 \circ f] \in \pi_1(A, x_0)$. Moreover, $[r_1 \circ f] = [f]$.

Then \(i_*([r_1 \circ f]) = [i \circ (r_1 \circ f)] = [r_1 \circ f] = [f]\), so \(i_*\) is surjective.

Hence, \(i_*\) is an isomorphism from $\pi_1(A, x_0)$ into $\pi_1(X, x_0)$.