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:
   • $\varphi_*([f]):\eq[\varphi\circ f]$

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 $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces, let $\varphi:X\rightarrow Y$ be a continuous map and let $p\in X$ serve as the base point 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

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

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
• $M_\varphi:\Omega(X,p)\rightarrow\Omega(Y,\varphi(p))$ by $M:f\mapsto(\varphi\circ f)$ is the core of the definition, the map taking loops to their images

In this case we claim that:

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

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

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

• factoring is in play.

Then we show it is a group homomorphism, ie:

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

Well-definedness of $\varphi_*$

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

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

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

Proof:

• Let $f,g\in\Omega(X,p)$ be given
  • Suppose that $\mathbb{P}_X(f)\neq\mathbb{P}_X(g)$ - then by the nature of logical implication we're done, we do not care about the truth or falsity of the RHS
  • Suppose that $\mathbb{P}_X(f)\eq\mathbb{P}_X(g)$ - then we must show that in this case $\mathbb{P}_Y(M_\varphi(f))\eq\mathbb{P}_Y(M_\varphi(g))$
    • We have $\mathbb{P}_X(f)\eq\mathbb{P}_X(g)$, that means:
      • $f\simeq g\ (\text{rel }\{0,1\})$
        • By the composition of end-point-preserving-homotopic paths with a continuous map yields end-point-preserving-homotopic paths, this means:
          • $(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})$
    • Thus $\mathbb{P}_X(f)\eq\mathbb{P}_X(g)\implies (\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})$
      • Furthermore $(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})$ is exactly the definition of $[\varphi\circ f]\eq[\varphi\circ g]$ (i.e. $\mathbb{P}_Y(\varphi\circ f)\eq\mathbb{P}_Y(\varphi\circ g)$) (which are obviously in $\pi_1(Y,\varphi(p))$ of course)
    • Next let us simplify the RHS:
      1. $\mathbb{P}_Y(M_\varphi(f))$

         $\eq\mathbb{P}_Y(\varphi\circ f)$

         $\eq[\varphi\circ f]$

      2. $\mathbb{P}_Y(M_\varphi(g))$

         $\eq\mathbb{P}_Y(\varphi\circ g)$

         $\eq[\varphi\circ g]$

    • We have already shown that we have $[\varphi\circ f]\eq[\varphi\circ g]$ from the LHS
  • So the entire thing boiled down to:
    • $\big[f\simeq g\ (\text{rel }\{0,1\})\big]\implies\big[(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})\big]$
• Since $f,g\in\Omega(X,p)$ we arbitrary we have shown it for all.

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

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

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.