Is There Only One Possible Z-Linear Map from T to T'?

[Math Amateur]

First, thanks to both Deveno and ThePerfectHacker for helping me to gain a basic understanding of tensor products of modules.

In a chat room discussion ThePerfectHacker suggested I show that \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.

I was aware of a solution to this exercise in Paul Garret's book Abstract Algebra - see Chapter 27 : Tensor Products, page 396 (see pages 395-396 attached) . However, I need help with Garrett's solution since I do not completely understand it ... so I am presenting his solution ... and my alternative solution ... I am hoping someone can indicate a correct solution ...

Note first that on page 395 in Section 27.3 First Examples (see attachment), Paul Garret writes:

" ... ... we emphasize that in \(\displaystyle M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N \), we can always rearrange

\(\displaystyle (rm) \otimes n = r(m \otimes n) = m \otimes (rn) \) ... ... ... (1)

Also for \(\displaystyle r, s \in R \),

\(\displaystyle (r + s)(m \otimes n) = rm \otimes n + sm \otimes n \) ... ... ... (2)

... ... "

but how or why these relations are true, I have very little idea ... can someone please help me to see why these would hold given the basic structure of the tensor product?Now on page 396 (Example 27.3.2 - see attachment) Garret gives the solution to the following exercise:

Show \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.

Garrett proceeds as follows:

For \(\displaystyle m \in {\mathbb{Z}}_a \text{ and } n \in {\mathbb{Z}}_b \) we have the following:

\(\displaystyle m \otimes n \)

\(\displaystyle = 1 \cdot (m \otimes n) \)

\(\displaystyle (ra + sb)(m \otimes n) \)

\(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) [here I think Garrett has a typo!]

Now at this point (correcting for the typo) Garrett seems to me to reason as follows:

\(\displaystyle ra(m \otimes n) = ar(m \otimes n) = a(rm \otimes n) \) by the commutativity of \(\displaystyle \mathbb{Z} \) and by (1) above.

Garrett then seems to reason as follows:

\(\displaystyle a(rm \otimes n) = a.0 = 0 \)

and also similarly

\(\displaystyle b(m \otimes sn) = b.0 = 0 \)

BUT why are \(\displaystyle (rm \otimes n) \) and \(\displaystyle (m \otimes sn) \) equal to zero? Can someone please help?

Now, since I did not understand Garrett's solution, I looked for another way to show what was required and proceeded as follows:

\(\displaystyle ra(m \otimes n) = r(am \otimes n) = r(0 \otimes n) \) since \(\displaystyle am = 0 \)

and then continue to reason that \(\displaystyle r(0 \otimes n) = r.0 = 0 \)

BUT ... I am really uncertain as to whether \(\displaystyle (0 \otimes n) = 0 \)

However, if I am correct then a similar chain of reasoning then proceeds and follows:

\(\displaystyle sb(m \otimes n) = s(m \otimes bn) = r(m \otimes 0) \) since \(\displaystyle bn = 0 \)

and so then \(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) = 0 + 0 = 0

Can someone please clarify the above for me?

Peter

 

[ThePerfectHacker]

Forget about tensor products for the time being. Do this exercise.

Let $M$ be a $\mathbb{Z}$-module.
Let $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ be a bilinear map.

Show that $f$ is the zero map i.e. $f(a,b) = 0$ for any $(a,b) \in \mathbb{Z}_n\times \mathbb{Z}_m$.

Hint: Choose integers $x,y$ so that $nx+my = 1$. Now,
$$ f(a,b) = f\big( 1\cdot (a,b) \big) = f\bigg( (nx+my)(a,b) \bigg) = .. $$

 

[Math Amateur]

Forget about tensor products for the time being. Do this exercise.

Let $M$ be a $\mathbb{Z}$-module.
Let $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ be a bilinear map.

Show that $f$ is the zero map i.e. $f(a,b) = 0$ for any $(a,b) \in \mathbb{Z}_n\times \mathbb{Z}_m$.

Hint: Choose integers $x,y$ so that $nx+my = 1$. Now,
$$ f(a,b) = f\big( 1\cdot (a,b) \big) = f\bigg( (nx+my)(a,b) \bigg) = .. $$

So trying your exercise ...

\(\displaystyle f(a,b) = f\big( 1\cdot (a,b) \big) = f\bigg( (nx+my)(a,b) \bigg) \)

\(\displaystyle = f\bigg( (nx + my)a, (nx+ my)b \bigg) \) [is this step valid?]

\(\displaystyle = f\bigg( nxa + mya, nxb+ myb \bigg) \)

where \(\displaystyle nxa + mya \in \mathbb{Z}_m \)

and \(\displaystyle nxb+ myb \in \mathbb{Z}_n \)

Now \(\displaystyle mya = 0 \text{ in } \mathbb{Z}_m \)

and \(\displaystyle nxb = 0\text{ in } \mathbb{Z}_n \)

SO we now have

\(\displaystyle f(a,b) = f(nxa, myb) \) ...

... but where to from here ...

Am I approaching this correctly?

Peter

EDIT - I know that I have not yet used bilinear properties of f yet ... but struggling to see how to use them profitably - could use properties

f(rm, n) = rf(m,n) and f(m, rn) = rf(m,n)

to bring some of the factors of nxa and myb out the "front" of f ... but ...

 

[ThePerfectHacker]

\(\displaystyle f(a,b) = f\big( 1\cdot (a,b) \big) = f\bigg( (nx+my)(a,b) \bigg) \)

\(\displaystyle = f\bigg( (nx + my)a, (nx+ my)b \bigg) \) [is this step valid?]

Yes it is valid because $k(a,b) = (a,b) + ... + (a,b) = (a+...+a,b+...+b) = (ka,kb)$. But this does not help you.

You need to use bilinear properties of $f$,
$$ f\bigg( (nx+my)(a,b) \bigg) = (nx+my)f(a,b) = nxf(a,b) + myf(a,b)$$

Now, $nxf(a,b) = f(nxa,b)$, why?, and as $nxa = 0$, why?, we get $nxf(a,b) = f(0,b)$.
In a similar way $myf(a,b) = f(a,0)$.

From here it follows that $f(0,b) = f(0\cdot 0,b) = 0f(0,b) = 0$. Similarly $f(a,0) = 0$.
Thus, $f(a,b) = 0 + 0 = 0$ which shows that $f$ is zero map.

 

[Math Amateur]

Yes it is valid because $k(a,b) = (a,b) + ... + (a,b) = (a+...+a,b+...+b) = (ka,kb)$. But this does not help you.

You need to use bilinear properties of $f$,
$$ f\bigg( (nx+my)(a,b) \bigg) = (nx+my)f(a,b) = nxf(a,b) + myf(a,b)$$

Now, $nxf(a,b) = f(nxa,b)$, why?, and as $nxa = 0$, why?, we get $nxf(a,b) = f(0,b)$.
In a similar way $myf(a,b) = f(a,0)$.

From here it follows that $f(0,b) = f(0\cdot 0,b) = 0f(0,b) = 0$. Similarly $f(a,0) = 0$.
Thus, $f(a,b) = 0 + 0 = 0$ which shows that $f$ is zero map.

Thanks ThePerfectHacker!

Just a couple of clarifications regarding

\(\displaystyle f\bigg( (nx+my)(a,b) \bigg) = (nx+my)f(a,b) = nxf(a,b) + myf(a,b) \)

What is the justification for

\(\displaystyle f\bigg( (nx+my)(a,b) \bigg) = (nx+my)f(a,b) \)

Why exactly does this follow?

I imagine that \(\displaystyle (nx+my)f(a,b) = nxf(a,b) + myf(a,b) \) folows by the distributivity of \(\displaystyle \mathbb{Z}\). Is that correct?You ask: \(\displaystyle nxf(a,b) = f(nxa,b) \), why? Answer bilinearity of f - specifically

f(rm, n) = rf(m, n)

You also ask - nxa = 0, why? - because nxa is in \(\displaystyle \mathbb{Z}_a \) and in that ring ma = 0 for any integer m and nx is an integer.Peter

 

[ThePerfectHacker]

What is the justification for

\(\displaystyle f\bigg( (nx+my)(a,b) \bigg) = (nx+my)f(a,b) \)

Why exactly does this follow?

That is correct, but for a silly reason. After all $nx+my = 1$, it is just that.

It is bad of me to write that equation because it gives the impression that it has something to do with bilinear properties. It does not. It is not a bilinear property that $f(k(a,b)) = kf(a,b)$. Only that $f(ka,b) = kf(a,b) = f(a,kb)$.

Okay now. Do you see why the tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is 0-module?

Theorem: Given two $R$-modules, $M$ and $N$, there exists an $R$-module, $T$, together with a bilinear map $b:M\times N \to T$ so that given any module $P$ and a bilinear map $f:M\times N\to P$ there exists a unique $R$-module homomorphism $\varphi:T\to P$ such that the diagram commutes. We call $T$ the tensor product of $M$ and $N$.

Can you solve the problem now?

 

[Math Amateur]

That is correct, but for a silly reason. After all $nx+my = 1$, it is just that.

It is bad of me to write that equation because it gives the impression that it has something to do with bilinear properties. It does not. It is not a bilinear property that $f(k(a,b)) = kf(a,b)$. Only that $f(ka,b) = kf(a,b) = f(a,kb)$.

Okay now. Do you see why the tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is 0-module?

Theorem: Given two $R$-modules, $M$ and $N$, there exists an $R$-module, $T$, together with a bilinear map $b:M\times N \to T$ so that given any module $P$ and a bilinear map $f:M\times N\to P$ there exists a unique $R$-module homomorphism $\varphi:T\to P$ such that the diagram commutes. We call $T$ the tensor product of $M$ and $N$.

Can you solve the problem now?

Thanks again ... still struggling to draw things together, but given that the tensor product of two R-modules is the result of a bilinear map \(\displaystyle f: M \times N \to M \times N \) (should I really be saying "the result of") , then there seems some sense in the conclusion that if the bilinear map is the zero map then the tensor product is the zero module.

Does that reasoning sound about right?

(I suspect I am being a bit vague and indulging in 'hand-waving" rather than a precise argument :()

I also suspect that there is some tie-up or relation between the statements by Garrett on page 395 that he uses to solve the problem/exercise, namely

" ... ... we emphasize that in \(\displaystyle M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N \), we can always rearrange

\(\displaystyle (rm) \otimes n = r(m \otimes n) = m \otimes (rn) \) ... ... ... (1)

Also for \(\displaystyle r, s \in R \),

\(\displaystyle (r + s)(m \otimes n) = rm \otimes n + sm \otimes n \) ... ... ... (2)

... ... "and the bilinearity of f. Is that correct?

Peter

 

[ThePerfectHacker]

Does that reasoning sound about right?

I do not know what you are sayin'.

Given two modules $M$ and $N$, there exists a module $T$, so that ... diagram commutes. This module $T$ is called the tensor product of $M$ and $N$.

To show that the tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is $0$-module you need to show that $T =0$. Show that $T$ satisfies the conditions of the theorem.

 

[Math Amateur]

I do not know what you are sayin'.

Given two modules $M$ and $N$, there exists a module $T$, so that ... diagram commutes. This module $T$ is called the tensor product of $M$ and $N$.

To show that the tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is $0$-module you need to show that $T =0$. Show that $T$ satisfies the conditions of the theorem.

Sorry ThePerfectHacker

What you say is correct ... we need to show T = 0 ..

So we need to relate our showing that the function f is the zero function with T being the zero module ... am I now making more sense?

I am now going back to your post with the Theorem in it to study the post and the theorem more carefully

Peter

 

[ThePerfectHacker]

Sorry ThePerfectHacker

What you say is correct ... we need to show T = 0 ..

So we need to relate our showing that the function f is the zero function with T being the zero module ... am I now making more sense?

Peter

The tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is a $\mathbb{Z}$-module $T$ for which there is a bilinear map $b:\mathbb{Z}_n\times \mathbb{Z}_m \to T$ so that given any module $M$ and bilinear map $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ there exists a homomorphism $\varphi:T\to M$ such that $f = \varphi b$.

You need to show that by taking $T=0$ you can find a bilinear map $b:\mathbb{Z}_n\times \mathbb{Z}_m \to 0 $ such that for any module $M$ and bilinear map $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ you can find a homomorphism $\varphi:0\to M$ so that $f = \varphi b$.

 

[Math Amateur]

The tensor product of $\mathbb{Z}_n$ and $\mathbb{Z}_m$ is a $\mathbb{Z}$-module $T$ for which there is a bilinear map $b:\mathbb{Z}_n\times \mathbb{Z}_m \to T$ so that given any module $M$ and bilinear map $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ there exists a homomorphism $\varphi:T\to M$ such that $f = \varphi b$.

You need to show that by taking $T=0$ you can find a bilinear map $b:\mathbb{Z}_n\times \mathbb{Z}_m \to 0 $ such that for any module $M$ and bilinear map $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ you can find a homomorphism $\varphi:0\to M$ so that $f = \varphi b$.

Just a slight clarification as I work through the above ...

You write:

"You need to show that by taking $T=0$ you can find a bilinear map ... ... "

Isn't taking \(\displaystyle T = 0\) assuming what we want to show ... ?
... Can you clarify?

Surely, what we want to show is that if $T=0$ exists then it must necessarily be zero? ... maybe I am misunderstanding something ... can you clarify?

Peter

 

[ThePerfectHacker]

Just a slight clarification as I work through the above ...

You write:

"You need to show that by taking $T=0$ you can find a bilinear map ... ... "

Isn't taking \(\displaystyle T = 0\) assuming what we want to show ... ?
... Can you clarify?

Peter

No it is not.

In the first exercise you showed that given any bilinear $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ it is the trivial zero map.

Now define $b:\mathbb{Z}_n\times \mathbb{Z}_m \to 0$ by $b(a,b) = 0$, this is trivially bilinear.

This module $T=0$ and the trivial bilinear map $b(a,b) = 0$ now satisfy the universal property condition, because, if $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ is bilinear we know it is the zero map (by exercise) and so if we let $\varphi: 0 \to M$ to trivial map, it follows that $f = \varphi b$. This shows that $T=0$ together with this trivial bilinear map $b$ satisfy the universal property of tensor product. Thus, $0$ is the tensor product between $\mathbb{Z}_n$ and $\mathbb{Z}_m$.

 

[Math Amateur]

No it is not.

In the first exercise you showed that given any bilinear $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ it is the trivial zero map.

Now define $b:\mathbb{Z}_n\times \mathbb{Z}_m \to 0$ by $b(a,b) = 0$, this is trivially bilinear.

This module $T=0$ and the trivial bilinear map $b(a,b) = 0$ now satisfy the universal property condition, because, if $f:\mathbb{Z}_n\times \mathbb{Z}_m \to M$ is bilinear we know it is the zero map (by exercise) and so if we let $\varphi: 0 \to M$ to trivial map, it follows that $f = \varphi b$. This shows that $T=0$ together with this trivial bilinear map $b$ satisfy the universal property of tensor product. Thus, $0$ is the tensor product between $\mathbb{Z}_n$ and $\mathbb{Z}_m$.

Thanks!

Just to be sure ...

When you write "the trivial bilinear map $b(a,b) = 0$" the bilinear map b is different from the element b in (a,b)? I guess it must be ...

Also ... just to clarify ... when you write;

" ... ... [FONT=&quot]so if we let [/FONT][FONT=&quot]φ [/FONT][FONT=&quot]:[/FONT] [FONT=&quot]0[/FONT] [FONT=&quot]→[/FONT] [FONT=&quot]M[/FONT][FONT=&quot][/FONT] to trivial map ... ... "

do you mean:

" ... ...[FONT=&quot]so if we let [/FONT][FONT=&quot]φ [/FONT][FONT=&quot]:[/FONT] [FONT=&quot]0[/FONT] [FONT=&quot]→[/FONT] [FONT=&quot]M[/FONT][FONT=&quot][/FONT] be the trivial map ... ... "

I am just clarifying not trying to be picky ...

Peter

 

[ThePerfectHacker]

Yes, $b(a,b) = 0$ means the map is $b$ and it sends the pair $(a,b)\to 0$. I know it is bad-notation but it works.

[One of my favorite double-notation abuses is when some people define a map $\pi:\mathbb{R}\to \mathbb{C}$ by $\pi(\theta) = e^{2\pi i \theta}$, clearly the $\pi$ for function and $\pi$ for number are clashing with one another!]

Yes, I meant to say " .. be the trivial map " instead of " .. to the trivial map".

 

[Deveno]

One can, in fact, prove (with not much more effort) that:

$\Bbb Z_a \otimes_{\Bbb Z} \Bbb Z_b \cong \Bbb Z_{\text{gcd}(a,b)}$

And then for $a,b$ co-prime we get that the tensor product is:

$\Bbb Z_1 = \Bbb Z/\Bbb Z \cong \{0\}$

using the fact that:

$\text{gcd}(a,b) = \min\{ra + sb \in \Bbb Z^+: r,s \in \Bbb Z\}$.

A word of caution on tensor products: the ring we tensor over makes a LOT of difference.

 

[ThePerfectHacker]

Here is another exercise.

Notation: Given two modules $M,N$ over a ring $R$ we denote $M\otimes N$ to be a module that satisfies the universal property i.e. there is a bilinear $b:M\times N\to M\otimes N$ so that given any module $P$ and bilinear $f:M\times N\to P$ there exists a unique homomorphism $\varphi:M\otimes N\to P$ such that $f=\varphi b$. Furthermore, for $m\in M$ and $n\in N$ we denote $m\otimes n \in M\otimes N$ to be $b(m,n)$.

Just a warning, not every element of $M\otimes N$ has the form $m\otimes n$, i.e. the blinear map $b:M\times N\to M\otimes N$ is not onto. Elements of $M\otimes N$ of the form $m\otimes n$ are called simple tensors. But not every element of the tensor product is a simple tensor.

Exercise: Prove that $(M_1\oplus M_2)\otimes N \simeq (M_1\otimes N)\oplus (M_2\otimes N)$.

 

[Math Amateur]

Yes, $b(a,b) = 0$ means the map is $b$ and it sends the pair $(a,b)\to 0$. I know it is bad-notation but it works.

[One of my favorite double-notation abuses is when some people define a map $\pi:\mathbb{R}\to \mathbb{C}$ by $\pi(\theta) = e^{2\pi i \theta}$, clearly the $\pi$ for function and $\pi$ for number are clashing with one another!]

Yes, I meant to say " .. be the trivial map " instead of " .. to the trivial map".

Hi ThePerfectHacker,

I notice you have set another problem on tensor products ... thank you, I will get onto the problem shortly ... I am, however, still reflecting on your solution to the previous exercise ... I had some trouble following the logic, so I will just sum up what I think has happened ... can you please confirm whether I am correct (or not) ... I will follow you very closely anyway, but please pick up any errors or misconceptions ...

By the way I am changing your notation a bit with the bilinear map b now being called \(\displaystyle \beta \) ... just for clarity ...
We have to show \(\displaystyle \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m = 0 \) where 0 is the zero \(\displaystyle \mathbb{Z}\)-module.
The definition of \(\displaystyle \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \) is given as follows:

----------------------------------------------------------------------------------------------

Definition. The tensor product of \(\displaystyle \mathbb{Z}_n \text{ and } \mathbb{Z}_m \) is a \(\displaystyle \mathbb{Z}\)-module, \(\displaystyle T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \)

together with a \(\displaystyle \mathbb{Z}\)-bilinear map \(\displaystyle \beta : \ \mathbb{Z}_n \times \mathbb{Z}_m \to T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \)

such that, for every \(\displaystyle \mathbb{Z}\)-bilinear map:

\(\displaystyle f : \ \mathbb{Z}_n \times \mathbb{Z}_m \to M \)

there exists a homomorphism (or actually just a linear map)

\(\displaystyle \phi : \ T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \to M \)

such that \(\displaystyle f = \phi \beta \)

-----------------------------------------------------------------------------------------------

Again, we have to show:

\(\displaystyle \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m = 0 \)

Following your posts, you are saying that we should put (or assume that) \(\displaystyle T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m = 0 \) and hence we have \(\displaystyle \beta : \ \mathbb{Z}_n \times \mathbb{Z}_m \to T = 0 \) i.e. \(\displaystyle \beta (a,b) = 0 \text{ for all} a \in \mathbb{Z}_n \text{ and } b \in \mathbb{Z}_m \).

Then we see from the definition above, that if T = 0 is actually the case then we will be able to find \(\displaystyle \phi : \ T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \to M \) such that \(\displaystyle f = \phi \beta \).Now we have shown in the previous exercise that every \(\displaystyle \mathbb{Z}\)-bilinear map \(\displaystyle f \) of the form

\(\displaystyle f : \ \mathbb{Z}_n \times \mathbb{Z}_m \to M \)

is the zero map \(\displaystyle f(a,b) = 0 \text{ for all} a \in \mathbb{Z}_n \text{ and } b \in \mathbb{Z}_m \)So, our solution to the current exercise is to let \(\displaystyle \phi : \ T = \mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Z}_m \to M \) be defined by \(\displaystyle \phi (a \otimes_\mathbb{Z} b ) = 0 \text{ for all } a \in \mathbb{Z}_n \text{ and } b \in \mathbb{Z}_m \) i.e. to let \(\displaystyle \phi \) be the zero map.

So now we have both \(\displaystyle \phi \) and \(\displaystyle \beta \) equal to the zero map - and the zero map composed with the zero map is the zero map! - so

\(\displaystyle f = \phi \beta \)

Thus we put \(\displaystyle T = 0 \) and found the unique (but is it unique? why?) bilinear map \(\displaystyle \phi \) such that \(\displaystyle f = \phi \beta \) and thus \(\displaystyle T = 0 \) is a solution.

But how do we know that there is not another T for which we can find a different \(\displaystyle \phi \) such that \(\displaystyle f = \phi \beta \) i.e. how do we know that our solution \(\displaystyle T = 0 \) is unique - that is the only T for which there is a \(\displaystyle \phi \) for which \(\displaystyle f = \phi \beta \).

Hoping someone can help confirm the above and help with my questions/issues.

Peter

 

[ThePerfectHacker]

Yes that is all correct.

The map $\varphi$ is unique because it is a homomorphism from $0$ to $M$, and there is only one such possible homomorphism, sending $0$ to $0$.

 

[Deveno]

Suppose there were another such, $T'$. What is the only possible $\Bbb Z$-linear map:

$T \to T'$ (what must any map that factors through a zero map be)?

Argue by symmetry that this map is bijective (it might help to draw the resulting diagrams).