View source for A continuous map induces a homomorphism on fundamental groups

{{Stub page|grade=A|msg=Important work!
# The definition is a little messy and due to the clutter doesn't show the most important part:
#* {{M|\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.
__TOC__
==Statement==
Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]], let {{M|\varphi:X\rightarrow Y}} be a [[continuous map]] and let {{M|p\in X}} serve as the base point for the [[fundamental group]] of {{M|X}} at {{M|p}}, {{M|\pi_1(X,p)}}. Then{{rITTMJML}}:
* {{M|\varphi_\ast:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))}} defined by {{M|\varphi_\ast:[f]\mapsto[\varphi\circ f]}} is a [[group homomorphism|homomorphism of the fundamental groups of {{M|X}} and {{M|Y}}]]
{{Caveat|We are implicitly claiming it is well defined:}} as we do not have {{M|f}} when we write {{M|[f]}}, to obtain {{M|f}} we must look at the inverse relation of the canonical projection, {{M|\mathbb{P}_X^{-1}([f]) }} in the notation developed next, giving us a set of all things equivalent to {{M|f}} and for any of these {{M|\varphi_\ast}} must yield the same result.
* {{M|\varphi_\ast}} is called the [[fundamental group homomorphism induced by a continuous map|homomorphism induced by the continuous map {{M|\varphi}}]]
===Formal definition===
{{Diagram|1=<m>\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))
}</m>
|2=test}}
With our situation we automatically have the following (which do not use their conventional symbols):
* {{M|\mathbb{P}_X:}}[[Omega(X,b)|{{M|\Omega(X,p)}}]]{{M|\rightarrow\pi_1(X,p)}}<ref group="Note">{{M|\pi_X}} is not used for the canonical projection because {{M|\pi}} is already in play as the fundamental group. Although it wouldn't lead to ambiguous writings, it's not helpful</ref><ref group="Note">Recall that [[Omega(X,b)|{{M|\Omega(X,p)}}]] is the [[set]] of all {{link|loop|topology|s}} in {{M|X}} based at {{M|p\in X}}. There is an operation, [[loop concatenation]], but it isn't a [[monoid]] or even a [[semigroup]] yet! As concatenation is not associative</ref> is the [[canonical projection of the equivalence relation]]
** i.e. {{M|\mathbb{P}_X:f\mapsto [f]\in\frac{\Omega(X,p)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})\big)} }};
* {{M|\mathbb{P}_Y:\Omega(Y,\varphi(p))\rightarrow\pi_1(Y,\varphi(p))}} is the canonical projection as above but for {{M|Y}}, and
* {{M|M_\varphi:\Omega(X,p)\rightarrow\Omega(Y,\varphi(p))}} by {{M|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{{rITTMJML}}:
* {{M|\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 [[fundamental group homomorphism induced by a continuous map|homomorphism induced by the continuous map {{M|\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:
* {{XXX|Expand on this, link to the identity property and the composition property on page 197.6 of{{rITTMJML}}}}
==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 {{M|[f]}} that there are many choices of map, say {{M|g}} such that {{M|[g]\eq [f]}}, we must check it is well defined, which as usual means:
* [[factor (function)|factoring]] is in play.
Then we show it is a group homomorphism, ie:
* {{M|1=\forall a,b\in\pi_1(X,p)[\varphi_*(a\cdot b)\eq\varphi_*(a)\cdot\varphi_*(b)]}}
===[[Well-definedness]] of {{M|\varphi_*}}===
{{Diagram|1=<m>\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))
}</m>
|2=test}}
We wish to {{link|factor|function}} to yield the map {{M|\overline{M_\varphi} }} as shown on the right. To apply the theorem we must first show:
* {{M|\forall f,g\in\Omega(X,p)\big[\big(\mathbb{P}_X(f)\eq\mathbb{P}_X(g)\big)\implies}} {{M|\big(\mathbb{P}_Y(M_\varphi(f))\eq\mathbb{P}_Y(M_\varphi(g))\big)\big]}}
'''Proof:'''
* Let {{M|f,g\in\Omega(X,p)}} be given
** Suppose that {{M|\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 {{M|\mathbb{P}_X(f)\eq\mathbb{P}_X(g)}} - then we must show that in this case {{M|\mathbb{P}_Y(M_\varphi(f))\eq\mathbb{P}_Y(M_\varphi(g))}}
*** We have {{M|\mathbb{P}_X(f)\eq\mathbb{P}_X(g)}}, that means:
**** {{M|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:
****** {{M|(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})}}
*** Thus {{M|\mathbb{P}_X(f)\eq\mathbb{P}_X(g)\implies (\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})}}
**** Furthermore {{M|(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})}} is exactly the definition of {{M|[\varphi\circ f]\eq[\varphi\circ g]}} (i.e. {{M|\mathbb{P}_Y(\varphi\circ f)\eq\mathbb{P}_Y(\varphi\circ g)}}) (which are obviously in {{M|\pi_1(Y,\varphi(p))}} of course)
*** Next let us simplify the RHS:
***# {{M|\mathbb{P}_Y(M_\varphi(f))}}
***#: {{M|\eq\mathbb{P}_Y(\varphi\circ f)}}
***#: {{M|\eq[\varphi\circ f]}}
***# {{M|\mathbb{P}_Y(M_\varphi(g))}}
***#: {{M|\eq\mathbb{P}_Y(\varphi\circ g)}}
***#: {{M|\eq[\varphi\circ g]}}
*** We have already shown that we have {{M|[\varphi\circ f]\eq[\varphi\circ g]}} from the LHS
** So the entire thing boiled down to:
*** [[the composition of end-point-preserving-homotopic paths with a continuous map yields end-point-preserving-homotopic paths|{{M|\big[f\simeq g\ (\text{rel }\{0,1\})\big]\implies\big[(\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})\big]}}]]
* Since {{M|f,g\in\Omega(X,p)}} we arbitrary we have shown it for all.

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

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

===[[Group homomorphism]]===
We want to show that:
* {{M|\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 {{M|[f],[g]\in\pi_1(X,p)}} be given.
** We now operate on the LHS and RHS:
**# The LHS:
**#* {{M|\varphi_\ast([f]\cdot[g])}}
**#*: {{M|\eq\varphi_\ast([f*g])}} (by the operation of the [[fundamental group]]) - note that {{M|*}} here denotes [[loop concatenation]] of course.
**#*: {{M|\eq[\varphi\circ(f*g)]}} (by definition of {{M|\varphi_\ast}})
**# The RHS:
**#* {{M|\varphi_\ast([f])\cdot\varphi_\ast([g])}}
**#*: {{M|\eq[\varphi\circ f]\cdot[\varphi\circ g]}}
**#*: {{M|\eq[(\varphi\circ f)*(\varphi\circ g)]}}
** Now we must show they're equal.
**# Using the definition of [[loop concatenation]] we see {{M|\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)}}{{M|\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.}}
**# Also using the definition of loop concatenation we see {{M|\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 {{M|[f],[g]\in\pi_1(X,p)}} were arbitrary we have shown this for all. As required.
{{Proofreading of proof required|grade=D|msg=Typos at most}}
==Notes==
<references group="Note"/>
==References==
<references/>
{{Theorem Of|Topology|Algebraic Topology|Homotopy Theory}}