Homotopic maps induce the same homomorphism
$begingroup$
This exercise is from Kosniowski's Algebraic Topology, page 137, exercise 15.11, d.
Show that two continuous maps $varphi, psi :X rightarrow Y$ with $varphi(x_0) = psi (x_0)$ for some $x_0 in X$ induce the same homomorphism from $pi(X,x_0)$ to $pi(Y, varphi(x_0))$ if $varphi$ and $psi$ are homotopic relative to $x_0$.
We need to show that, given the equivalence class of a loop with base $x_0$, $[g]in pi(X,x_0)$, then $[varphi circ g ] = [psi circ g ]$.
The fact that $varphi(x_0) = psi (x_0)$ means that $[varphi circ g]$ and $[psi circ g]$ are both equivalence classes of loops in $Y$ with base point $varphi(x_0)$. But I don't see how the fact that $varphi$ is homotopic to $psi$ implies that $[varphi circ g]=[psi circ g]$.
algebraic-topology
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
$endgroup$
add a comment |
$begingroup$
This exercise is from Kosniowski's Algebraic Topology, page 137, exercise 15.11, d.
Show that two continuous maps $varphi, psi :X rightarrow Y$ with $varphi(x_0) = psi (x_0)$ for some $x_0 in X$ induce the same homomorphism from $pi(X,x_0)$ to $pi(Y, varphi(x_0))$ if $varphi$ and $psi$ are homotopic relative to $x_0$.
We need to show that, given the equivalence class of a loop with base $x_0$, $[g]in pi(X,x_0)$, then $[varphi circ g ] = [psi circ g ]$.
The fact that $varphi(x_0) = psi (x_0)$ means that $[varphi circ g]$ and $[psi circ g]$ are both equivalence classes of loops in $Y$ with base point $varphi(x_0)$. But I don't see how the fact that $varphi$ is homotopic to $psi$ implies that $[varphi circ g]=[psi circ g]$.
algebraic-topology
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
$endgroup$
add a comment |
$begingroup$
This exercise is from Kosniowski's Algebraic Topology, page 137, exercise 15.11, d.
Show that two continuous maps $varphi, psi :X rightarrow Y$ with $varphi(x_0) = psi (x_0)$ for some $x_0 in X$ induce the same homomorphism from $pi(X,x_0)$ to $pi(Y, varphi(x_0))$ if $varphi$ and $psi$ are homotopic relative to $x_0$.
We need to show that, given the equivalence class of a loop with base $x_0$, $[g]in pi(X,x_0)$, then $[varphi circ g ] = [psi circ g ]$.
The fact that $varphi(x_0) = psi (x_0)$ means that $[varphi circ g]$ and $[psi circ g]$ are both equivalence classes of loops in $Y$ with base point $varphi(x_0)$. But I don't see how the fact that $varphi$ is homotopic to $psi$ implies that $[varphi circ g]=[psi circ g]$.
algebraic-topology
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
$endgroup$
This exercise is from Kosniowski's Algebraic Topology, page 137, exercise 15.11, d.
Show that two continuous maps $varphi, psi :X rightarrow Y$ with $varphi(x_0) = psi (x_0)$ for some $x_0 in X$ induce the same homomorphism from $pi(X,x_0)$ to $pi(Y, varphi(x_0))$ if $varphi$ and $psi$ are homotopic relative to $x_0$.
We need to show that, given the equivalence class of a loop with base $x_0$, $[g]in pi(X,x_0)$, then $[varphi circ g ] = [psi circ g ]$.
The fact that $varphi(x_0) = psi (x_0)$ means that $[varphi circ g]$ and $[psi circ g]$ are both equivalence classes of loops in $Y$ with base point $varphi(x_0)$. But I don't see how the fact that $varphi$ is homotopic to $psi$ implies that $[varphi circ g]=[psi circ g]$.
algebraic-topology
algebraic-topology
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
asked Dec 13 '18 at 17:32
YaggerYagger
8581517
8581517
add a comment |
add a comment |
1 Answer
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$begingroup$
You can prove this more general result:
Lemma. Let $Asubset X,$ $Bsubset Y.$ If $f_0,f_1:Xto Y$ and $g_0,g_1:Yto Z$ are maps, $f_0simeq f_1$ rel $A,$ $g_0simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)subset B,$ then $g_0circ f_0simeq g_1circ f_1$ rel $A.$
Proof sketch. If $h:Xtimes Ito Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Ytimes Ito Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:Xtimes Ito Z$ given by $$H(x,t)=h'(h(x,t),t)$$
gives the desired homotopy from $g_0circ f_0$ to $g_1circ f_1$ rel $A.$
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can prove this more general result:
Lemma. Let $Asubset X,$ $Bsubset Y.$ If $f_0,f_1:Xto Y$ and $g_0,g_1:Yto Z$ are maps, $f_0simeq f_1$ rel $A,$ $g_0simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)subset B,$ then $g_0circ f_0simeq g_1circ f_1$ rel $A.$
Proof sketch. If $h:Xtimes Ito Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Ytimes Ito Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:Xtimes Ito Z$ given by $$H(x,t)=h'(h(x,t),t)$$
gives the desired homotopy from $g_0circ f_0$ to $g_1circ f_1$ rel $A.$
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
$begingroup$
You can prove this more general result:
Lemma. Let $Asubset X,$ $Bsubset Y.$ If $f_0,f_1:Xto Y$ and $g_0,g_1:Yto Z$ are maps, $f_0simeq f_1$ rel $A,$ $g_0simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)subset B,$ then $g_0circ f_0simeq g_1circ f_1$ rel $A.$
Proof sketch. If $h:Xtimes Ito Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Ytimes Ito Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:Xtimes Ito Z$ given by $$H(x,t)=h'(h(x,t),t)$$
gives the desired homotopy from $g_0circ f_0$ to $g_1circ f_1$ rel $A.$
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
$begingroup$
You can prove this more general result:
Lemma. Let $Asubset X,$ $Bsubset Y.$ If $f_0,f_1:Xto Y$ and $g_0,g_1:Yto Z$ are maps, $f_0simeq f_1$ rel $A,$ $g_0simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)subset B,$ then $g_0circ f_0simeq g_1circ f_1$ rel $A.$
Proof sketch. If $h:Xtimes Ito Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Ytimes Ito Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:Xtimes Ito Z$ given by $$H(x,t)=h'(h(x,t),t)$$
gives the desired homotopy from $g_0circ f_0$ to $g_1circ f_1$ rel $A.$
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
$endgroup$
You can prove this more general result:
Lemma. Let $Asubset X,$ $Bsubset Y.$ If $f_0,f_1:Xto Y$ and $g_0,g_1:Yto Z$ are maps, $f_0simeq f_1$ rel $A,$ $g_0simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)subset B,$ then $g_0circ f_0simeq g_1circ f_1$ rel $A.$
Proof sketch. If $h:Xtimes Ito Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Ytimes Ito Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:Xtimes Ito Z$ given by $$H(x,t)=h'(h(x,t),t)$$
gives the desired homotopy from $g_0circ f_0$ to $g_1circ f_1$ rel $A.$
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 17:42
positrón0802positrón0802
4,468520
4,468520
add a comment |
add a comment |
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