A continuous map induces a homomorphism on fundamental groups

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Important work!

Statement

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces, let $\varphi:X\rightarrow Y$ be a continuous map and let $p\in X$ be the base point we consider for the fundamental group of $X$ at $p$, $\pi_1(X,p)$. Then:

• $\varphi_\ast:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))$ defined by $\varphi_\ast:[f]\mapsto[\varphi\circ f]$ is a homomorphism of the fundamental groups of $X$ and $Y$

Caveat:We are implicitly claiming it is well defined: as we do not have $f$ when we write $[f]$, to obtain $f$ we must look at the inverse relation of the canonical projection, $\mathbb{P}_X^{-1}([f])$ in the notation developed next, giving us a set of all things equivalent to $f$ and for any of these $\varphi_\ast$ must yield the same result.

• $\varphi_\ast$ is called the homomorphism induced by the continuous map $\varphi$

Formal definition

$\xymatrix{ \Omega(X,p) & & \Omega(Y,\varphi(p)) \\ \pi_1(X,p) & & \pi_1(Y,\varphi(p)) }$
Diagram

With our situation we automatically have the following (which do not use their conventional symbols):

• $\mathbb{P}_X:$$\Omega(X,p)$$\rightarrow\pi_1(X,p)$ is the canonical projection of the equivalence relation, i.e. $\mathbb{P}_X:f\mapsto [f]\in\frac{\Omega(X,p)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})\big)}$;
• $\mathbb{P}_Y:\Omega(Y,\varphi(p))\rightarrow\pi_1(Y,\varphi(p))$ is the canonical projection as above but for $Y$, and
• $M:\Omega(X,p)\rightarrow\Omega(Y,\varphi(p))$ by $M:f\mapsto(\varphi\circ f)$

In this case we claim that:

• $\varphi_\ast(\alpha):\eq\mathbb{P}_Y(M(\mathbb{P}^{-1}_X(\alpha)))$ is an unambiguous (i.e. is well-defined) definition and is a group homomorphism.
  • It is called the homomorphism induced by the continuous map $\varphi$

OLD Statement

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces, let $\varphi:X\rightarrow Y$ be a continuous map and let $p\in X$ be the base point we consider for the fundamental group of $X$ at $p$, $\pi_1(X,p)$ then we also have the following:

• $\mathbb{P}_X:$$\Omega(X,p)$$\rightarrow\pi_1(X,p)$ is the canonical projection of the equivalence relation, i.e. $\mathbb{P}_X:f\mapsto [f]\in\frac{\Omega(X,p)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})\big)}$;
• $\mathbb{P}_Y:\Omega(Y,\varphi(p))\rightarrow\pi_1(Y,\varphi(p))$ is the canonical projection as above but for $Y$, and
• $M:\Omega(X,p)\rightarrow\Omega(Y,\varphi(p))$ by $M:f\mapsto(\varphi\circ f)$

In this case we claim that:

• $\varphi_\ast(\alpha):\eq\mathbb{P}_Y(M(\mathbb{P}^{-1}_X(\alpha)))$ is an unambiguous (i.e. is well-defined) definition and is a group homomorphism.
  • It is called the homomorphism induced by the continuous map $\varphi$

Informal definition

Informally, we define $\varphi_\ast$ as follows:

• $\varphi_\ast:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))$ defined by $\varphi_\ast:[f]\mapsto[\varphi\circ f]$ is a group homomorphism

We claim that this is well-defined and that it is indeed a group homomorphism

Proof

Well-definedness of $\varphi_*$

The notation $\varphi_*([f])$ makes it seem like $f$ is some given loop. Remember that we're actually dealing with equivalence classes not a loop, thus:

• for $\alpha\in\pi_1(X,p)$ we must define $\varphi_*(\alpha)$ - not so obvious now! We actually define:
  • {{M|\varphi_*(\alpha):\eq [f\circ\mathbb{P}^{-1}_X

Group homomorphism

We want to show that:

• $\forall [f],[g]\in\pi_1(X,p)\big[\varphi_\ast([f]\cdot[g])\eq\varphi_\ast([f])\cdot\varphi_\ast([g])\big]$

We will do this by operating on the left-hand-side (LHS) and the right-hand-side (RHS) separately.

• Let $[f],[g]\in\pi_1(X,p)$ be given.
  • We now operate on the LHS and RHS:
    1. The LHS:
       • $\varphi_\ast([f]\cdot[g])$

         $\eq\varphi_\ast([f*g])$ (by the operation of the fundamental group) - note that $*$ here denotes loop concatenation of course.

         $\eq[\varphi\circ(f*g)]$ (by definition of $\varphi_\ast$)

    2. The RHS:
       • $\varphi_\ast([f])\cdot\varphi_\ast([g])$

         $\eq[\varphi\circ f]\cdot[\varphi\circ g]$

         $\eq[(\varphi\circ f)*(\varphi\circ g)]$

  • Now we must show they're equal.
    1. Using the definition of loop concatenation we see $\text{LHS}\eq\varphi\circ\left(\left\{\begin{array}{lr}f(2t)&\text{for }t\in[0,\frac{1}{2}]\\ g(2t-1)&\text{for }t\in[\frac{1}{2},1]\end{array}\right.\right)$$\eq\left\{\begin{array}{lr}\varphi(f(2t))&\text{for }t\in[0,\frac{1}{2}]\\\varphi(g(2t-1)&\text{for }t\in[\frac{1}{2},1]\end{array}\right.$
    2. Also using the definition of loop concatenation we see $\text{RHS}\eq\left\{\begin{array}{lr}\varphi(f(2t))&\text{for }t\in[0,\frac{1}{2}]\\\varphi(g(25-1))&\text{for }t\in[\frac{1}{2},1]\end{array}\right.$
  • Clearly these are the same
• Since $[f],[g]\in\pi_1(X,p)$ were arbitrary we have shown this for all. As required.

Stuff

By the composition of end-point-preserving-homotopic paths with a continuous map yields end-point-preserving-homotopic paths we know that if:

• $f_1,f_2:I\rightarrow X$ are paths and $\varphi$ is a continuous map, as stated above, that:
  • If $f_1\simeq f_2\ (\text{rel }\{0,1\})$ then $(\varphi\circ f_1)\simeq(\varphi\circ f_2)\ (\text{rel }\{0,1\})$