Hatcher dump
by yayups, Sep 27, 2020, 8:15 PM
Section 1.1
As a general convention,
will be the path variable and 
will be the homotopy variable.
As a general convention,
will be the path variable and 
will be the homotopy variable.1.1.1 wrote:
Show that the composition of paths satisfies the following cancellation property: If 
and 
, then 
.
and 
, then 
.1.1.2 wrote:
Show that the change-of-basepoint homomorphism 
depends only on the homotopy class of 
.
depends only on the homotopy class of 
.
,![\begin{align*} \beta_h:\pi_1(X,x_1)&\to \pi_1(X,x_0) \\ [f]&\mapsto [h\cdot f\cdot \overline{h}] = [h]\cdot[f]\cdot[h]^{-1}. \end{align*}](http://latex.artofproblemsolving.com/4/7/4/4747e0b0715c43482201733e2800fb679404f357.png)
This only depends on ![$[h]$](http://latex.artofproblemsolving.com/5/5/8/5584515cc49cb2b664ff811b0d429147eb37f5df.png)
,1.1.3 wrote:
For a path connected space 
, show that 
is abelian iff all basepoint-change homomorphisms depend only on the endpoints of the path .
, show that 
is abelian iff all basepoint-change homomorphisms depend only on the endpoints of the path .![\begin{align*} \beta_h:\pi_1(X,x_1)&\to \pi_1(X,x_0) \\ [f]&\mapsto [h]\cdot[f]\cdot[h]^{-1}. \end{align*}](http://latex.artofproblemsolving.com/d/6/2/d62f57d28d8ca576663e71e544eeebdef8a8b997.png)
Fix any paths 
from 
from 
.![\begin{align*} \beta_{\bar{h_1}} = \beta_{h_2} &\iff [h_1]^{-1}\cdot[f]\cdot[h_1] = [h_2]^{-1}\cdot[f]\cdot[h_2]\quad\forall f\in\pi_1(X,x_1) \\ &\iff [f]\cdot [g] = [g]\cdot [f], \end{align*}](http://latex.artofproblemsolving.com/f/5/6/f56227e4d15f571660dd985a73c49f38cb80e58f.png)
and since we can attain any 
this way, we're done.1.1.4 wrote:
A subspace 
is said to be star-shaped if there is a point 
such that, for each 
, the line segment from 
to 
lies in . Show that if a subspace 
is locally star-shaped, in the sense that every point of has a star-shaped neighborhood in , then every path in is homotopic to a piecewise linear path, that is, a path consisting of a finite number of straight line segments traversed at constant speed. Show this applies in particular when is open or when is a union of finitely many closed convex sets.
is said to be star-shaped if there is a point 
such that, for each 
, the line segment from 
to 
lies in . Show that if a subspace 
is locally star-shaped, in the sense that every point of has a star-shaped neighborhood in , then every path in is homotopic to a piecewise linear path, that is, a path consisting of a finite number of straight line segments traversed at constant speed. Show this applies in particular when is open or when is a union of finitely many closed convex sets.
is star-shaped. Then, any path 
is homotopic to a piecewise linear path with at most 
pieces. Note that in this homotopy of paths, the endpoints remain fixed.![$[-1,1]\to U$](http://latex.artofproblemsolving.com/9/5/8/9583814894ff6a632453e34f20c50fa241f08f87.png)
.
be the center of the star of 
.![\[F(s,t) = [1-s(1-|t|)]\cdot g(t).\]](http://latex.artofproblemsolving.com/e/b/1/eb1d3c0917fe9722394ec9560c703352a44c0a5a.png)
Note this stays in 
.
to a composition of two paths, one from some point to ![\[g:I\to U\]](http://latex.artofproblemsolving.com/e/f/8/ef87fb8aed5c13ae5bd14e49c5c978135965d7a6.png)
such that 
.![\[F(s,t) = t\cdot g(s+(1-s)t),\]](http://latex.artofproblemsolving.com/d/d/b/ddb2827345060efad0efd136b65d087f0b121f94.png)
(which is a homotopy for the same reason) which sends 
.
![\[f:I\to X,\]](http://latex.artofproblemsolving.com/2/2/7/227fbfbb6ab5e9d445902e771fa4f01e45f0be4e.png)
where 
denote the open neighborhood of 
is an open set in 
,
,
an open interval that contains 
of them such that the 
cover 
s
that is between them and contained in 
.![\[0=t_0<t_1<t_2<\cdots<t_{m'}=1,\]](http://latex.artofproblemsolving.com/3/6/7/367d9537badc5068d533c5cf884fe853df18638e.png)
such that ![$[t_i,t_{i+1}]$](http://latex.artofproblemsolving.com/9/a/b/9aba5be221b8c6726708a61dabf49e970e6978f1.png)
is fully contained in some 
,
pieces.1.1.5 wrote:
Show that for a space , the following three conditions are equivalent:
is simply connected iff all maps 
are homotopic.
, the following three conditions are equivalent:
- Every map is homotopic to a constant map, with image a point.
- Every map extends to a map
. 
for all .
.
for all is simply connected iff all maps 
are homotopic.
has a single component. We'll view 
.
be a path from 
,
in the following way: ![\[[\tilde{f}] = [\eta_{f(1)}]\cdot[g]\cdot[\eta_{f(1)}],\]](http://latex.artofproblemsolving.com/7/1/7/717be21cf5f988b83b241743129940e14fbf9c9d.png)
where 
.
,
by 
.
,
is continuous where 
is the quotient map).
(
be a homotopy such that 
and 
(constant). Now, define 
by ![\[\tilde{F}(t,s) = \widetilde{F(\bullet,s)}(t).\]](http://latex.artofproblemsolving.com/0/e/2/0e29b63964a7144648ad77fe157c1a67333abdee.png)
We see that this a homotopy from ![$[\eta_{x_1}]\cdot[\eta_{x_1}]^{-1}$](http://latex.artofproblemsolving.com/6/9/3/693f7ba6b948d0364cb25b3dfe0a1281e761ba22.png)
,
with 
and 
.
given by ![\[\tilde{F}(\omega,s) = \widetilde{F(\bullet,s)}(\omega)\]](http://latex.artofproblemsolving.com/1/3/0/13096561d44b117daf8dfc9b50c5f717437dfc2a.png)
This is a homotopy from 
(
is homeomorphic to 
where 
is identified to a point, and furthermore, this homeomorphism identically maps 
to the boundary of ![\[F:S^1\times I/\sim\,\,\to X\]](http://latex.artofproblemsolving.com/0/1/d/01dc8001c9fa32e7d8304db406fadd08fc82ca28.png)
such that 
for a fixed map ![\[\tilde{F}:S^1\times I\to X\]](http://latex.artofproblemsolving.com/7/0/9/709f0ba9ad799e29fd4679f282e97bace67fbd27.png)
such that 
and 
is constant. This is exactly equivalent to saying that 1.1.6 wrote:
We can regard 
as the set of basepoint-preserving homotopy classes of maps 
. Let ![$[S^1,X]$](//latex.artofproblemsolving.com/c/1/1/c11e4e8e083dbe4a02d442d3e41f63e74e5e3fdf.png)
be the set of homotopy classes of maps with no conditions on basepoints. Thus there is a natural map ![$\Phi:\pi_1(X,x_0)\to [S^1,X]$](//latex.artofproblemsolving.com/5/0/0/500056a0ab5daca828eda0bfbe701170c74eb16a.png)
obtained by ignoring basepoints. Show that 
is onto if is path connected, and that ![$\Phi([f])=\Phi([g])$](//latex.artofproblemsolving.com/2/a/3/2a3aef364a8d2ce1e27513617b70bab92925a3ff.png)
iff ![$[f]$](//latex.artofproblemsolving.com/f/4/4/f44bfb37b68791e7968284f5b7237bb59749656a.png)
and ![$[g]$](//latex.artofproblemsolving.com/f/e/6/fe6f53448f516922ffc8265065784c6aa7db8525.png)
are conjugate in . Hence induces a one-to-one correspondence between and the set of conjugacy classes in , when is path connected.
as the set of basepoint-preserving homotopy classes of maps 
. Let ![$[S^1,X]$](http://latex.artofproblemsolving.com/c/1/1/c11e4e8e083dbe4a02d442d3e41f63e74e5e3fdf.png)
be the set of homotopy classes of maps with no conditions on basepoints. Thus there is a natural map ![$\Phi:\pi_1(X,x_0)\to [S^1,X]$](http://latex.artofproblemsolving.com/5/0/0/500056a0ab5daca828eda0bfbe701170c74eb16a.png)
obtained by ignoring basepoints. Show that 
is onto if is path connected, and that ![$\Phi([f])=\Phi([g])$](http://latex.artofproblemsolving.com/2/a/3/2a3aef364a8d2ce1e27513617b70bab92925a3ff.png)
iff ![$[f]$](http://latex.artofproblemsolving.com/f/4/4/f44bfb37b68791e7968284f5b7237bb59749656a.png)
and ![$[g]$](http://latex.artofproblemsolving.com/f/e/6/fe6f53448f516922ffc8265065784c6aa7db8525.png)
are conjugate in . Hence induces a one-to-one correspondence between and the set of conjugacy classes in , when is path connected.
and 
notation from the previous solution. For any ![$\Phi([a]\cdot[\tilde{f}]\cdot[a]^{-1})$](http://latex.artofproblemsolving.com/b/a/7/ba7d0501fc86af993ad0d629057ad5f5227d7c96.png)
is homotopic to 
is a path from 
,![$\Phi([f]\cdot [g]) = \Phi([g]\cdot [f])$](http://latex.artofproblemsolving.com/4/c/4/4c44b623825d730a4bab5fbe921071c778bbedb6.png)
for all 
,![$\Phi([f]\cdot[g])$](http://latex.artofproblemsolving.com/f/6/b/f6bf1cbb3bcff3bb71af57deb4bf9134dac67a28.png)
with a rotation to get ![$\Phi([g]\cdot[f])$](http://latex.artofproblemsolving.com/6/2/9/62996a727e65b2c6fb3b0ad43c9ee7b28e25c2e1.png)
.![$\Phi([f])\to\Phi([g])$](http://latex.artofproblemsolving.com/9/c/f/9cfcc27492cf550020f60f14cab830f1686d8a59.png)
be ![$g([0,t])$](http://latex.artofproblemsolving.com/3/5/5/3556ba065dabc6d1e47a98716ae36d835b51e932.png)
for some 
,
,![$[g]=[g_2]\cdot[g_1]$](http://latex.artofproblemsolving.com/e/3/c/e3cba19e3466b3db94314bbc887d0cadc8f2bc4b.png)
.![\[[a]\cdot[g_1]\cdot[g_2]\cdot[a]^{-1} = [f].\]](http://latex.artofproblemsolving.com/8/1/8/8181ee80c51213d2453b40acd190a6030dfcb519.png)
It is not hard to check that this rearranges to ![\[[g] = ([a]\cdot[g_1])^{-1}\cdot[f]\cdot([a]\cdot[g_1]),\]](http://latex.artofproblemsolving.com/a/8/8/a88c060d9a54353a1a5f1c73214892d4f6f326a2.png)
and since ![$[a]\cdot[g_1]\in\pi_1(X,x_0)$](http://latex.artofproblemsolving.com/e/9/8/e985e079ccbbe1ad18f2ea53a0ba81c28a7a0c65.png)
,1.1.7 wrote:
Define 
by 
, so 
restricts to the identity on the two boundary circles of 
. Show that is homotopic to the identity by a homotopy that is stationary on one of the boundary circles, but not by any homotopy that is stationary on both.
by 
, so 
restricts to the identity on the two boundary circles of 
. Show that is homotopic to the identity by a homotopy that is stationary on one of the boundary circles, but not by any homotopy that is stationary on both.
works just fine.
(here 
given by gluing the two boundary circles together. Since 
is stationary on the boundary circles, it is a continuous map from 
.
,
.This post has been edited 2 times. Last edited by yayups, Sep 27, 2020, 8:18 PM
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are the integers because the interval retracts to a point. The image of 
April 2021
September 2020