MATH 422 Lecture Note #3 (2018 Spring)Induced homomorphisms and products

Jae Choon Cha
POSTECH $\def\id{\operatorname{id}}\def\rel{\text{ rel }}\def\R{\mathbb{R}}\def\C{\mathbb{C}}$

In the above, we discussed fundamental groups, as algebraic objects obtained from spaces. Now we construct group homomorphisms from continuous maps.

Induced homomorphisms

The idea is natural. Recall that an element of the fundamental group of a space $X$ is represented by a loop, which is a map $\alpha\colon I\to X$. For a map $X\to Y$, by composing it with $\alpha$, we obtain a new loop in $Y$, which corresponds to the given loop $\alpha$ in $X$. In what follows, we will show that this association gives rise to a well-defined homomorphism.

Definition. Suppose $f\colon X\to Y$ is a map, $x_0\in X$, and $y_0=f(x_0)$. Define $f_*\colon \pi_1(X,x_0) \to \pi_1(Y,y_0)$ by $f_*([\alpha]) = [f\circ \alpha]$.

Lemma. $f_*$ is a well-defined group homomorphism.

Proof. Since compositions of homotopic maps are homotopic, we have $f\circ\alpha\simeq f\circ\beta \rel\{0,1\}$ whenever $\alpha\simeq \beta\rel\{0,1\}$. From this it follows that $f_*$ is well-defined. Since $f\circ(\alpha* \beta) = (f\circ \alpha)* (f\circ\beta)$, $f_*$ is a homomorphism.

We call $f_*$ the homomorphism induced by $f$.

Homotopy and induced homomorphisms

The following theorem describes the relationship of the homomorphisms induced by homotopic maps. Roughly speaking, we want to say that homotopic maps induce the same homomorphisms, but there is a technicality. In general, since a homotopy does not preserve the basepoint, homotopic maps may induce homomorphisms to different groups. This issue is resolved as in the following statement.

Theorem. Suppose $F\colon f\simeq g$ is a homotopy from $f$ to $g\colon X\to Y$, and $x_0\in X$. Define a path $\alpha\colon I\to Y$ by $\alpha(s)=F(x_0,s)$. Then $\alpha_* \circ g_* = f_*$. That is, the following diagram is commutative.

running…

Here, note that $\alpha_*$ is the basepoint change isomorphism associated to $\alpha$; since $\alpha$ is a path from $F(x_0,0)=f(x_0)$ to $F(x_0,1)=g(x_1)$, $\alpha_*$ is from $\pi_1(Y,g(x_0))$ to $\pi_1(Y,f(x_0))$.

Proof of Theorem. Suppose $\gamma$ is a loop in $X$ based at $x_0$, i.e., $[\gamma] \in \pi_1(X,x_0)$. Define $G\colon I\times I \to Y$ by $G(s,t)=F(\gamma(s),t)$. Observe that $G(s,0)=(f\circ\gamma)(s)$, $G(s,1)=(g\circ\gamma)(s)$ and $G(0,s) = G(1,s)=\alpha(s)$. See the diagram below.

running…

Now, by the standard argument used earlier, which hinges on the convexity of $I\times I$, it follows that $\alpha(g\circ\gamma)\alpha^{-1} \simeq f\circ\gamma \rel\{0,1\}$. Since $[\alpha(g\circ\gamma)\alpha^{-1}] = \alpha_ *(g_ *([\gamma]))$ and $[f\circ\gamma]=f_*([\gamma])$, this completes the proof. $\quad\square$

Corollary. If two maps $f$, $g\colon X\to Y$ are homotopic rel $\{x_0\}$, then $f_* =g_*$.

Proof. In this case, $f(x_0)=g(x_0)$, and the path $\alpha$ in the above theorem is the constant path $c_{f(x_0)}$, which induces the identity on $\pi_1(Y,f(x_0))$. So, the corollary immediately follows from the theorem. $\quad\square$

Funtoriality

The following two facts look very innocent, but they are of great importance.

Theorem. (1) $(\id_X)_* = \id_{\pi_1(X,x_0)}$ (2) $(g\circ f)_* = g_*\circ f_*$.

Note that (1) holds for any $x_0\in X$ and (2) holds whenever $g\circ f$ is defined, for any choice of a basepoint of the domain of $f$.

We remark that when an association (which is $f \mapsto f_*$ together with $(X,x_0) \mapsto \pi_1(X,x_0)$ in our case) satisfies (1) and (2), then we often say that it is functorial (or natural). Functoriality is observed in many cases. Indeed, it is the fundamental idea in the abstract notion of categories and functors, which you will learn later (unfornatunately not in this course). Though, I strongly recommend you to remember that the properties (1) and (2) will be seen repeatedly in your future journey in mathematics.

Since the proof of the above theorem is straightforward, we leave it to the readers. Indeed, a standard use of functoriality which is illustrated below is more important.

Theorem. If $f\colon X\to Y$ is a homotopy equivalence, $f_*\colon \pi_1(X,x_0)\to \pi_1(Y,f(x_0))$ is an isomorphism for any $x_0\in X$.

Proof. Let $g\colon Y\to X$ be a homotopy inverse of $g$. Fix $x_0\in X$. SInce $g\circ f\simeq \id_X$, $g_* \circ f_* = \alpha_* \circ \id_{\pi_1(X,x_0)}= \alpha_*$ for some path $\alpha$ in $X$. Since $\alpha_*$ is onto and 1-1, $g_*$ is onto and $f_*$ is 1-1. Exchanging the roles of $f$ and $g$, we obtain that $f_*$ is onto and $g_*$ is 1-1. Consequently, both $f_*$ and $g_*$ are bijective. $\quad\square$

This tells us that homotopy equivalent spaces have the same fundamental group, up to isomorphisms.

Fundamental group of a product space

As an application of induced maps, we will investigate the fundamental group of a product space. Recall that for two groups $G$ and $H$, their product $G\times H$ is defined to be the cartesian product $G\times H=\{(g,h)\mid g\in G,\, h\in H\}$ with the group operation $(g,h)\cdot (g',h') = (gg',hh')$. Also, If $\phi\colon \Gamma\to G$ and $\psi\colon \Gamma\to H$ are group homomorphisms, then the function $\phi\times\psi \colon \Gamma\to G\times H$ defined by $(\phi\times\psi)(a)=(\phi(a),\psi(a))$ is a group homomorphism.

Theorem. Let $p\colon X\times Y \to X$ be the projection $p(x,y)=x$, and let $q\colon X\times Y \to Y$ be $q(x,y)=y$. Then the homomorphism

$p_* \times q_* \colon \pi_1(X\times Y,(x_0,y_0)) \to \pi_1(X,x_0)\times \pi_1(Y,y_0)$
is an isomorphism.

Proof. For given $[\beta]\in \pi_1(X,x_0)$ and $[\gamma]\in \pi_1(Y,y_0)$, define $\alpha\colon I\to X\times Y$ by $\alpha(s)=(\beta(s),\gamma(s))$. Then $\alpha$ is a loop based at $(x_0,y_0)$, and so $[\alpha]\in \pi_1(X\times Y,(x_0,y_0))$. Since $p_* ([\alpha])=[p\circ\alpha]=[\beta]$ and $q_* ([\alpha])=[q\circ\alpha]=[\gamma]$, we have $(p_* \times q_* )([\alpha])=([\beta],[\gamma])$. This shows that $p_* \times q_*$ is surjective.

For the injectivity, suppose $[\alpha]$ is in the kernel of $p_* \times q_*$. It follows that $[p\circ\alpha] = p_* [\alpha] = [c_{x_0}]$ and similarly $[q\circ \alpha] = [c_{y_0}]$. So, there are homotopies $F\colon p\circ\alpha\simeq c_{x_0} \rel\{0,1\}$ and $G\colon q\circ\alpha \simeq c_{y_0}\rel\{0,1\}$. Define $H\colon I\times I \to X\times Y$ by $H(s,t)=(F(s,t),G(s,t))$. Then, it is straightforward to verify that $H$ is a homotopy $\alpha \simeq c_{(x_0,y_0)} \rel\{0,1\}$. That is, $[\alpha]$ is equal to the identity in the group $\pi_1(X\times Y,(x_0,y_0))$. This shows that $p_* \times q_*$ has trivial kernel. $\quad\square$