First de Rham Cohomology group of the 2-torus
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Show that for the de Rham cohomology, $H^{1}_{dR}({T^{2}})$ is isomorphic to $mathbb{R}^{2}$ by showing that the following map: $[alpha] to (int_{S^{1}}f^{*}_{1}alpha,int_{S^{1}}f^{*}_{1}alpha)$, where $f_{1}=(theta,c_{1}), f_{2}=(c_{2},theta)$ for $c_{1},c_{2}$ constants (seen as maps $S^{1} to S^{1} times S^{1}$) is an isomorphism.
I have shown that the map is well-defined, linear and independent of the constants chosen. However, I'm having trouble with showing that it is surjective and injective. Surjectivity seems to be easy to show, but I'm not sure on how to go about it. For injectivity, it suffices to show that if the integrals are $0$, then $alpha$ is an exact form, yet I'm having troubles with it as well. What would be a good way to tackle this problem?
differential-geometry algebraic-topology smooth-manifolds differential-forms
add a comment |
up vote
3
down vote
favorite
Show that for the de Rham cohomology, $H^{1}_{dR}({T^{2}})$ is isomorphic to $mathbb{R}^{2}$ by showing that the following map: $[alpha] to (int_{S^{1}}f^{*}_{1}alpha,int_{S^{1}}f^{*}_{1}alpha)$, where $f_{1}=(theta,c_{1}), f_{2}=(c_{2},theta)$ for $c_{1},c_{2}$ constants (seen as maps $S^{1} to S^{1} times S^{1}$) is an isomorphism.
I have shown that the map is well-defined, linear and independent of the constants chosen. However, I'm having trouble with showing that it is surjective and injective. Surjectivity seems to be easy to show, but I'm not sure on how to go about it. For injectivity, it suffices to show that if the integrals are $0$, then $alpha$ is an exact form, yet I'm having troubles with it as well. What would be a good way to tackle this problem?
differential-geometry algebraic-topology smooth-manifolds differential-forms
1
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Show that for the de Rham cohomology, $H^{1}_{dR}({T^{2}})$ is isomorphic to $mathbb{R}^{2}$ by showing that the following map: $[alpha] to (int_{S^{1}}f^{*}_{1}alpha,int_{S^{1}}f^{*}_{1}alpha)$, where $f_{1}=(theta,c_{1}), f_{2}=(c_{2},theta)$ for $c_{1},c_{2}$ constants (seen as maps $S^{1} to S^{1} times S^{1}$) is an isomorphism.
I have shown that the map is well-defined, linear and independent of the constants chosen. However, I'm having trouble with showing that it is surjective and injective. Surjectivity seems to be easy to show, but I'm not sure on how to go about it. For injectivity, it suffices to show that if the integrals are $0$, then $alpha$ is an exact form, yet I'm having troubles with it as well. What would be a good way to tackle this problem?
differential-geometry algebraic-topology smooth-manifolds differential-forms
Show that for the de Rham cohomology, $H^{1}_{dR}({T^{2}})$ is isomorphic to $mathbb{R}^{2}$ by showing that the following map: $[alpha] to (int_{S^{1}}f^{*}_{1}alpha,int_{S^{1}}f^{*}_{1}alpha)$, where $f_{1}=(theta,c_{1}), f_{2}=(c_{2},theta)$ for $c_{1},c_{2}$ constants (seen as maps $S^{1} to S^{1} times S^{1}$) is an isomorphism.
I have shown that the map is well-defined, linear and independent of the constants chosen. However, I'm having trouble with showing that it is surjective and injective. Surjectivity seems to be easy to show, but I'm not sure on how to go about it. For injectivity, it suffices to show that if the integrals are $0$, then $alpha$ is an exact form, yet I'm having troubles with it as well. What would be a good way to tackle this problem?
differential-geometry algebraic-topology smooth-manifolds differential-forms
differential-geometry algebraic-topology smooth-manifolds differential-forms
edited Nov 18 at 14:48
asked Nov 18 at 14:36
betelgeuse
305
305
1
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33
add a comment |
1
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33
1
1
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Recall that $H_1(T^2)$ is $mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $Omega^*(T^2) otimes C_*(T^2,mathbb R)tomathbb R$ such that $langle omega,sigmarangle = int_sigma omega$. This is the same as the corresponding map $Omega^*(T^2)to C^*(T^2,mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.
Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes
$$require{AMScd}
begin{CD} 0@>>>Omega^*(X) @>>> Omega^*(U)oplus Omega^*(V) @>>> Omega(Ucap V) @>>> 0\ {} @VVr_XV @VV{r_Uoplus r_V} V @VV{r_{Ucap V}}V\
0 @>>> C^*(X) @>>> C^*(U)oplus C^*(V) @>>> C^*(Ucap V) @>>> 0 end{CD}
$$
It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $Ucap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.
Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which
are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.
In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $dtheta$,
and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1to S^1$. Then the induced by on homology is simply the isomorphism $mathbb Rotimesmathbb Rtomathbb R$,
so you get what you wanted.
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall that $H_1(T^2)$ is $mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $Omega^*(T^2) otimes C_*(T^2,mathbb R)tomathbb R$ such that $langle omega,sigmarangle = int_sigma omega$. This is the same as the corresponding map $Omega^*(T^2)to C^*(T^2,mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.
Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes
$$require{AMScd}
begin{CD} 0@>>>Omega^*(X) @>>> Omega^*(U)oplus Omega^*(V) @>>> Omega(Ucap V) @>>> 0\ {} @VVr_XV @VV{r_Uoplus r_V} V @VV{r_{Ucap V}}V\
0 @>>> C^*(X) @>>> C^*(U)oplus C^*(V) @>>> C^*(Ucap V) @>>> 0 end{CD}
$$
It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $Ucap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.
Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which
are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.
In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $dtheta$,
and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1to S^1$. Then the induced by on homology is simply the isomorphism $mathbb Rotimesmathbb Rtomathbb R$,
so you get what you wanted.
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
add a comment |
up vote
2
down vote
accepted
Recall that $H_1(T^2)$ is $mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $Omega^*(T^2) otimes C_*(T^2,mathbb R)tomathbb R$ such that $langle omega,sigmarangle = int_sigma omega$. This is the same as the corresponding map $Omega^*(T^2)to C^*(T^2,mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.
Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes
$$require{AMScd}
begin{CD} 0@>>>Omega^*(X) @>>> Omega^*(U)oplus Omega^*(V) @>>> Omega(Ucap V) @>>> 0\ {} @VVr_XV @VV{r_Uoplus r_V} V @VV{r_{Ucap V}}V\
0 @>>> C^*(X) @>>> C^*(U)oplus C^*(V) @>>> C^*(Ucap V) @>>> 0 end{CD}
$$
It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $Ucap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.
Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which
are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.
In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $dtheta$,
and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1to S^1$. Then the induced by on homology is simply the isomorphism $mathbb Rotimesmathbb Rtomathbb R$,
so you get what you wanted.
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall that $H_1(T^2)$ is $mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $Omega^*(T^2) otimes C_*(T^2,mathbb R)tomathbb R$ such that $langle omega,sigmarangle = int_sigma omega$. This is the same as the corresponding map $Omega^*(T^2)to C^*(T^2,mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.
Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes
$$require{AMScd}
begin{CD} 0@>>>Omega^*(X) @>>> Omega^*(U)oplus Omega^*(V) @>>> Omega(Ucap V) @>>> 0\ {} @VVr_XV @VV{r_Uoplus r_V} V @VV{r_{Ucap V}}V\
0 @>>> C^*(X) @>>> C^*(U)oplus C^*(V) @>>> C^*(Ucap V) @>>> 0 end{CD}
$$
It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $Ucap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.
Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which
are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.
In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $dtheta$,
and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1to S^1$. Then the induced by on homology is simply the isomorphism $mathbb Rotimesmathbb Rtomathbb R$,
so you get what you wanted.
Recall that $H_1(T^2)$ is $mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $Omega^*(T^2) otimes C_*(T^2,mathbb R)tomathbb R$ such that $langle omega,sigmarangle = int_sigma omega$. This is the same as the corresponding map $Omega^*(T^2)to C^*(T^2,mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.
Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes
$$require{AMScd}
begin{CD} 0@>>>Omega^*(X) @>>> Omega^*(U)oplus Omega^*(V) @>>> Omega(Ucap V) @>>> 0\ {} @VVr_XV @VV{r_Uoplus r_V} V @VV{r_{Ucap V}}V\
0 @>>> C^*(X) @>>> C^*(U)oplus C^*(V) @>>> C^*(Ucap V) @>>> 0 end{CD}
$$
It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $Ucap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.
Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which
are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.
In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $dtheta$,
and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1to S^1$. Then the induced by on homology is simply the isomorphism $mathbb Rotimesmathbb Rtomathbb R$,
so you get what you wanted.
edited Nov 21 at 0:49
answered Nov 21 at 0:02
Pedro Tamaroff♦
95.7k10150295
95.7k10150295
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
add a comment |
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
This whole story is wonderfully explained in Bott and Tu's Differential forms in algebraic topoogy.
– Pedro Tamaroff♦
Nov 21 at 0:04
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
I see, we haven't covered Mayer-Vietoris sequences yet but this makes a lot of sense now.
– betelgeuse
Dec 2 at 14:06
add a comment |
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1
Do you already know the map $H^1_{dR}(S^1)rightarrow mathbb{R}$ given by $[alpha]mapsto int_{S^1} alpha$ is an isomorphism? If not, I would start there.
– Jason DeVito
Nov 18 at 16:56
Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}_{dR}(T^{2}) to mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F_{1}^{*}$ by $f_{1}^{*}$ with the isomorphism $H^{1}_{dR}(S^{1}) to mathbb{R}$ be a good way to proceed?
– betelgeuse
Nov 18 at 21:28
Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)in mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)rightarrow mathbb{R}$, there is a $1$-form $beta$ on $S^1$ with $int_{S^1} beta = r$. Now, let $alpha(theta_1, theta_2) = beta(theta_1)$ . Then $[alpha]mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$....
– Jason DeVito
Nov 21 at 16:32
use the fact that $H^1_{dR})(T^2)subseteq mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below.
– Jason DeVito
Nov 21 at 16:33