A continuous map induces a homomorphism on fundamental groups

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Important work!

Statement

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

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

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

• [ilmath]\varphi_\ast[/ilmath] is called the homomorphism induced by the continuous map [ilmath]\varphi[/ilmath]

Formal definition

[ilmath]\xymatrix{ \Omega(X,p) & & \Omega(Y,\varphi(p)) \\ \pi_1(X,p) & & \pi_1(Y,\varphi(p)) }[/ilmath]
Diagram

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

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

In this case we claim that:

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

OLD Statement

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

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

In this case we claim that:

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

Informal definition

Informally, we define [ilmath]\varphi_\ast[/ilmath] as follows:

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

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

Proof

Well-definedness of [ilmath]\varphi_*[/ilmath]

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

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

Group homomorphism

We want to show that:

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

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

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

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

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

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

         [ilmath]\eq[\varphi\circ f]\cdot[\varphi\circ g][/ilmath]

         [ilmath]\eq[(\varphi\circ f)*(\varphi\circ g)][/ilmath]

  • Now we must show they're equal.
    1. Using the definition of loop concatenation we see [ilmath]\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)[/ilmath][ilmath]\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.[/ilmath]
    2. Also using the definition of loop concatenation we see [ilmath]\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.[/ilmath]
  • Clearly these are the same
• Since [ilmath][f],[g]\in\pi_1(X,p)[/ilmath] 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:

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