A continuous map induces a homomorphism on fundamental groups

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Important work!
1. The definition is a little messy and due to the clutter doesn't show the most important part:
• [ilmath]\varphi_*([f]):\eq[\varphi\circ f][/ilmath]
Note: this page is the proof of the fundamental group homomorphism induced by a continuous map being 1) a map, and 2) a group homomorphism. See that page for the definition.

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] serve as the base point for the fundamental group of [ilmath]X[/ilmath] at [ilmath]p[/ilmath], [ilmath]\pi_1(X,p)[/ilmath]. Then[1]:

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.

Formal definition

 test [ilmath]\xymatrix{ \Omega(X,p) \ar[rr]^-{M_\varphi} \ar[d]_{\mathbb{P}_X} & & \Omega(Y,\varphi(p)) \ar[d]^{\mathbb{P}_Y} \\ \pi_1(X,p) \ar@{.>}[rr]_-{\varphi_*:\eq\overline{M_\varphi} } & & \pi_1(Y,\varphi(p)) }[/ilmath]

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][Note 1][Note 2] 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_\varphi:\Omega(X,p)\rightarrow\Omega(Y,\varphi(p))[/ilmath] by [ilmath]M:f\mapsto(\varphi\circ f)[/ilmath] is the core of the definition, the map taking loops to their images

In this case we claim that[1]:

See next

It would be better to look at the fundamental group homomorphism induced by a continuous map page, this is just a proof, but if not:

• TODO: Expand on this, link to the identity property and the composition property on page 197.6 of[1]

Proof

We use the context of the formal definition. Before we show "the map" is a group homomorphism we must show it actually is a map, as it is defined.

Remember that when we write [ilmath][f][/ilmath] that there are many choices of map, say [ilmath]g[/ilmath] such that [ilmath][g]\eq [f][/ilmath], we must check it is well defined, which as usual means:

Then we show it is a group homomorphism, ie:

• [ilmath]\forall a,b\in\pi_1(X,p)[\varphi_*(a\cdot b)\eq\varphi_*(a)\cdot\varphi_*(b)][/ilmath]

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

 test [ilmath]\xymatrix{ \Omega(X,p) \ar[rr]^-{M_\varphi} \ar[d]_{\mathbb{P}_X} & & \Omega(Y,\varphi(p)) \ar[d]^{\mathbb{P}_Y} \\ \pi_1(X,p) \ar@{.>}[rr]_-{\varphi_*:\eq\overline{M_\varphi} } & & \pi_1(Y,\varphi(p)) }[/ilmath]

We wish to factor to yield the map [ilmath]\overline{M_\varphi} [/ilmath] as shown on the right. To apply the theorem we must first show:

• [ilmath]\forall f,g\in\Omega(X,p)\big[\big(\mathbb{P}_X(f)\eq\mathbb{P}_X(g)\big)\implies[/ilmath] [ilmath]\big(\mathbb{P}_Y(M_\varphi(f))\eq\mathbb{P}_Y(M_\varphi(g))\big)\big][/ilmath]

Proof:

• Let [ilmath]f,g\in\Omega(X,p)[/ilmath] be given
• Suppose that [ilmath]\mathbb{P}_X(f)\neq\mathbb{P}_X(g)[/ilmath] - then by the nature of logical implication we're done, we do not care about the truth or falsity of the RHS
• Suppose that [ilmath]\mathbb{P}_X(f)\eq\mathbb{P}_X(g)[/ilmath] - then we must show that in this case [ilmath]\mathbb{P}_Y(M_\varphi(f))\eq\mathbb{P}_Y(M_\varphi(g))[/ilmath]
• We have [ilmath]\mathbb{P}_X(f)\eq\mathbb{P}_X(g)[/ilmath], that means:
• Thus [ilmath]\mathbb{P}_X(f)\eq\mathbb{P}_X(g)\implies (\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})[/ilmath]
• Furthermore [ilmath](\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})[/ilmath] is exactly the definition of [ilmath][\varphi\circ f]\eq[\varphi\circ g][/ilmath] (i.e. [ilmath]\mathbb{P}_Y(\varphi\circ f)\eq\mathbb{P}_Y(\varphi\circ g)[/ilmath]) (which are obviously in [ilmath]\pi_1(Y,\varphi(p))[/ilmath] of course)
• Next let us simplify the RHS:
1. [ilmath]\mathbb{P}_Y(M_\varphi(f))[/ilmath]
[ilmath]\eq\mathbb{P}_Y(\varphi\circ f)[/ilmath]
[ilmath]\eq[\varphi\circ f][/ilmath]
2. [ilmath]\mathbb{P}_Y(M_\varphi(g))[/ilmath]
[ilmath]\eq\mathbb{P}_Y(\varphi\circ g)[/ilmath]
[ilmath]\eq[\varphi\circ g][/ilmath]
• We have already shown that we have [ilmath][\varphi\circ f]\eq[\varphi\circ g][/ilmath] from the LHS
• So the entire thing boiled down to:
• Since [ilmath]f,g\in\Omega(X,p)[/ilmath] we arbitrary we have shown it for all.

Thus we may define [ilmath]\overline{M_\varphi}:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))[/ilmath] unambiguously by [ilmath]\overline{M_\varphi}:\alpha\mapsto\mathbb{P}_Y(M_\varphi(\mathbb{P}^{-1}_X(\alpha)))[/ilmath] - it doesn't matter which representative of [ilmath]\mathbb{P}^{-1}_X(\alpha)[/ilmath] we take.

This justifies the notation [ilmath]\overline{M_\varphi}:[f]\mapsto [\varphi\circ f][/ilmath] - as it doesn't matter whuch [ilmath]f[/ilmath] we take to represent the [[equivalence class] [ilmath][f][/ilmath].

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.

Notes

1. [ilmath]\pi_X[/ilmath] is not used for the canonical projection because [ilmath]\pi[/ilmath] is already in play as the fundamental group. Although it wouldn't lead to ambiguous writings, it's not helpful
2. Recall that [ilmath]\Omega(X,p)[/ilmath] is the set of all loops in [ilmath]X[/ilmath] based at [ilmath]p\in X[/ilmath]. There is an operation, loop concatenation, but it isn't a monoid or even a semigroup yet! As concatenation is not associative