The Induced Homo. from the Fund. Groups of Two Topo. Spaces

# The Induced Homomorphism from the Fundamental Groups of Two Topological Spaces

Definition: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a continuous function with $f(x) = y$. The Induced Homomorphism from $f$ of $\pi_1(X, x)$ to $\pi_1(Y, f(x))$ is the function $f_* : \pi_1(X, x) \to \pi_1(Y, f(x))$ defined for all homotopy classes $[\alpha] \in \pi_1(X, x)$ by $f_*([\alpha]) = [f \circ \alpha]$. |

*Observe that since $\alpha : I \to X$ is a loop in $X$ and $f$ is a continuous function, $f \circ \alpha : I \to Y$ is a loop in $Y$.*

We following theorem verifies that the induced homomorphism is indeed a homomorphism.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a continuous function with $f(x) = y$. Then the induced mapping $f_* : \pi_1(X, x) \to \pi_2(X, f(x))$ is a group homomorphism. |

**Proof:**Let $[\alpha], [\beta] \in \pi_1(X, x)$. Then:

\begin{align} \quad f_*([\alpha][\beta]) = f_*([\alpha\beta]) = [f \circ \alpha \beta] = [(f \circ \alpha)(f \circ \beta)] = [f \circ \alpha][f \circ \beta] = f_*([\alpha])f_*([\beta]) \end{align}

- Since:

\begin{align} \quad f \circ (\alpha \beta) = f \circ \left\{\begin{matrix} \alpha(2t) & 0 \leq t \leq \frac{1}{2}\\ \beta(2t - 1) & \frac{1}{2} \leq t \leq 1 \end{matrix}\right. = \left\{\begin{matrix} f \circ \alpha(2t) & 0 \leq t \leq \frac{1}{2}\\ f \circ\beta(2t - 1) & \frac{1}{2} \leq t \leq 1 \end{matrix}\right. = (f \circ \alpha)(f \circ \beta) \end{align}

- So indeed, $f_*$ is a homomorphism. $\blacksquare$

Theorem 2: Let $X$, $Y$, and $Z$ be topological spaces and let $f : X \to Y$ and $g : Y \to Z$ be continuous functions. Then $(g \circ f)_* = g_* \circ f_*$. |

**Proof:**Let $[\alpha] \in \pi_1(X, x)$. Then:

\begin{align} \quad g_* \circ f_* ([\alpha]) = g_* ([f \circ \alpha]) = [g \circ (f \circ \alpha)] = [(g \circ f) \circ \alpha] = (g \circ f)_*([\alpha]) \quad \blacksquare \end{align}

Theorem 3: Let $X$ be a topological space and let $A \subset X$ be a topological subspace. If $A$ is a retract of $X$ with retraction $r : X \to A$ then $r_* : \pi_1(X, a) \to \pi_1(A, a)$ is surjective. |

**Proof:**By theorem 2:

\begin{align} \quad r_* \circ \mathrm{in}_* = \mathrm{id}_* \end{align}

- But $\mathrm{id}_*$ is bijective. So the outer function, $r_*$ is surjective. $\blacksquare$

Corollary 4: Let $X$ be a topological space and let $A \subset X$ be a topological subspace. If $|\pi_1(A, a)| > |\pi_1(X, a)|$ then $A$ is not a retract of $X$. |

- If such a retraction $r : X \to A$ exists, then $r_* : \pi_1(X, a) \to \pi_1(A, a)$ is a surjection. But this cannot happen since the codomain is strictly larger than the domain. So $A$ is not a retract of $X$. $\blacksquare$

Theorem 5: Let $X$ be a topological space and let $A \subset X$ be a topological subspace. If $X$ is simply connected and $A$ is not simply connected then $A$ is not a retract of $X$. |

- Since $X$ is simply connected, $\pi_1(X, a) = \{ [c_a] \}$, the trivial group. And since $A$ is not simply connected, $\pi_2(A, a) \neq \{ [c_a] \}$, is not the trivial group. Suppose that $A$ is a retract of $X$. Then there exists a retraction $r : X \to A$. But by Theorem 3 we have that $r_* : \pi_1(X, a) \to \pi_1(A, a)$ is surjective. But $\pi_1(X, a)$ contains a single element while $\pi_1(A, a)$ constants at least two elements. So $r$ cannot be surjective which is a contradiction.

- So the assumption that $A$ is a retract of $X$ is false. So $A$ is NOT a retract of $X$. $\blacksquare$

Corollary 6: The unit circle $S'$ is not a retract of the closed unit disk $D^2$. |

**Proof:**The fundamental group of the unit circle is $\pi_1(S', \vec{x}) \cong \mathbb{Z}$ which is not simply connected, and the fundamental group of the closed unit disk is $\pi_1(D^2, \vec{x}) = \{ [c_{\vec{x}}] \}$ since $D^2$ is a convex subset of $\mathbb{R}^2$, which is simply connected. By Theorem 4, $S'$ is not a retract of $D^2$. $\blacksquare$

Theorem 7: Let $X$ be a topological space and let $A \subseteq X$ be a topological subspace. If $A$ is a deformation retract of $X$ then $\pi_1(X, x)$ is isomorphic to $\pi_1(A, a)$. |