Difference between revisions of "A continuous map induces a homomorphism on fundamental groups"
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− | {{Stub page|grade=A|msg=Important work!}} | + | {{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__ | __TOC__ | ||
==Statement== | ==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}} | + | 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}}]] | * {{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. | {{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 [[homomorphism induced by a continuous map|homomorphism induced by the continuous map {{M|\varphi}}]] | + | * {{M|\varphi_\ast}} is called the [[fundamental group homomorphism induced by a continuous map|homomorphism induced by the continuous map {{M|\varphi}}]] |
===Formal definition=== | ===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} \\ | \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)) | \pi_1(X,p) \ar@{.>}[rr]_-{\varphi_*:\eq\overline{M_\varphi} } & & \pi_1(Y,\varphi(p)) | ||
− | }</m | + | }</m> |
− | | | + | |2=test}} |
− | + | ||
− | + | ||
− | + | ||
With our situation we automatically have the following (which do not use their conventional symbols): | 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]] | * {{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]] | ||
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In this case we claim that{{rITTMJML}}: | 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]]. | * {{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 [[homomorphism induced by a continuous map|homomorphism induced by the continuous map {{M|\varphi}}]] | + | ** 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|\mathbb{P} | + | * {{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{ | |
− | * {{M|\ | + | \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}} | |
− | * {{M|\ | + | 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]]=== | |
− | + | ||
− | + | ||
− | ===Group homomorphism=== | + | |
We want to show that: | 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]}} | * {{M|\forall [f],[g]\in\pi_1(X,p)\big[\varphi_\ast([f]\cdot[g])\eq\varphi_\ast([f])\cdot\varphi_\ast([g])\big]}} | ||
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** Clearly these are the same | ** Clearly these are the same | ||
* Since {{M|[f],[g]\in\pi_1(X,p)}} were arbitrary we have shown this for all. As required. | * 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== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> |
Latest revision as of 07:35, 14 December 2016
- 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.
Contents
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]:
- [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
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]:
- [ilmath]\varphi_\ast(\alpha):\eq\mathbb{P}_Y(M_\varphi(\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]
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:
- factoring is in play.
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]
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:
- [ilmath]f\simeq g\ (\text{rel }\{0,1\})[/ilmath]
- By the composition of end-point-preserving-homotopic paths with a continuous map yields end-point-preserving-homotopic paths, this means:
- [ilmath](\varphi\circ f)\simeq(\varphi\circ g)\ (\text{rel }\{0,1\})[/ilmath]
- By the composition of end-point-preserving-homotopic paths with a continuous map yields end-point-preserving-homotopic paths, this means:
- [ilmath]f\simeq g\ (\text{rel }\{0,1\})[/ilmath]
- 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:
- [ilmath]\mathbb{P}_Y(M_\varphi(f))[/ilmath]
- [ilmath]\eq\mathbb{P}_Y(\varphi\circ f)[/ilmath]
- [ilmath]\eq[\varphi\circ f][/ilmath]
- [ilmath]\mathbb{P}_Y(M_\varphi(g))[/ilmath]
- [ilmath]\eq\mathbb{P}_Y(\varphi\circ g)[/ilmath]
- [ilmath]\eq[\varphi\circ g][/ilmath]
- [ilmath]\mathbb{P}_Y(M_\varphi(f))[/ilmath]
- We have already shown that we have [ilmath][\varphi\circ f]\eq[\varphi\circ g][/ilmath] from the LHS
- We have [ilmath]\mathbb{P}_X(f)\eq\mathbb{P}_X(g)[/ilmath], that means:
- 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:
- 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])
- [ilmath]\varphi_\ast([f]\cdot[g])[/ilmath]
- 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]
- [ilmath]\varphi_\ast([f])\cdot\varphi_\ast([g])[/ilmath]
- The LHS:
- Now we must show they're equal.
- 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]
- 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
- We now operate on the LHS and RHS:
- Since [ilmath][f],[g]\in\pi_1(X,p)[/ilmath] were arbitrary we have shown this for all. As required.
Template:Proofreading of proof required
Notes
- ↑ [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
- ↑ 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
References
- Stub pages
- XXX Todo
- Theorems
- Theorems, lemmas and corollaries
- Topology Theorems
- Topology Theorems, lemmas and corollaries
- Topology
- Algebraic Topology Theorems
- Algebraic Topology Theorems, lemmas and corollaries
- Algebraic Topology
- Homotopy Theory Theorems
- Homotopy Theory Theorems, lemmas and corollaries
- Homotopy Theory