Cartesian sum of subspace and quotient space isomorphic to whole space

[Eclair_de_XII]

Homework Statement

Let $X$ be a vector space and $Y$ a linear subspace. Prove that the Cartesian sum $Y\oplus X/Y$ is isomorphic to $X$.

Relevant Equations

Isomorphic:
Two spaces A and B are isomorphic to each other if there is a one-to-one and onto function f defined from A to B or vice versa.

Quotient space:
$X/Y=\{x\in X:\exists\,x_0 \in X \textrm{ such that } x-x_0=y\textrm{ for some }y\in Y\}$

Cartesian sum:
Let $A,B$ be two sets. Then the Cartesian sum of those sets is denoted as $A\oplus B$ and comprises of elements of the form $(a,b)$ where $a\in A,b\in B$. Addition and multiplication are defined component-wise.

Theorem:
$\dim Y\oplus {X/Y}=\dim X$

Let $n=\dim X$ and $m=\dim Y$.

Define a basis for $X: y_1,...,y_m,z_{m+1},...,z_n$. The first $m$ terms are a basis for $Y$. The remaining $n-m$ terms are a basis for its complement w.r.t $X$. Let's call it $Z$. $X$ is the direct sum of $Y$ and $Z$; denote it as $X=Y+Z$. In other words, you can express any $x\in X$ uniquely as $x=y+z$ for some $y\in Y,z\in Z$.

Define a linear transformation $T:X\rightarrow (Y\oplus X/Y)$ by $T(x)=(y,z)$ where $x=y+z$. We prove that it is one-to-one and unique.

Suppose for some $x_1\in X$, $T(x_1)=T(x)$. Then $(y_1,z_1)=(y,z)$ and in turn, $(y_1-y,z_1-z)=(0,0)$; $y_1=y$ and $z_1=z$ as a result. It follows that $x_1=y_1+z_1=y+z=x$.

Let $(y_0,z_0)\in Y\oplus Z$. Since $y_0\in Y$ and $z_0\in Z$, then $y_0+z_0\in Y+Z=X$. Hence, there is a $x_0\in X$ such that $x_0=y_0+z_0$. $T(x_0)=(y_0,z_0)$ as a result.

Hence, $Y\oplus X/Y$ is isomorphic to $X$.

 

[jbunniii]

Your definition of quotient space doesn't look right.

For one thing, every $x \in X$ is trivially in $\{x \in X : \exists x_0 \in X \text{ such that }x - x_0 = y \text{ for some }y \in Y\}$: just take $x_0 = x$ and $y = 0$.

More importantly, the elements of $X/Y$ are subsets of $X$ (namely, cosets of $Y$), not elements of $X$.

The correct definition is: $X / Y = \{x + Y : x \in X\}$, where $x + Y = \{x + y : y \in Y\}$.

 

[wrobel]

It must be a general fact I guess. Independently on finite or infinite dimensions. There is a subspace $W\subset X$ such that $X=Y\oplus W$. Let $p:X\to X/Y$ stand for the natural projection. Then $p\mid_W:W\to X/Y$ is an isomorphism. Is not it?

 

[fresh_42]

The statement doesn't mention finite dimensions!

You can work with a basis in finite dimensions.
$$
\underbrace{\underbrace{y_1,\ldots,y_m}_{\text{basis of }Y},x_{m+1},\ldots,x_n}_{\text{basis of }X}
$$
Now $n-m=\dim X/Y=\dim U$ where $U:=\operatorname{span}\{x_{m+1},\ldots,x_n\}$. Since all vector spaces of the same dimension are isomorphic, we get $X=Y\oplus U \cong X\oplus X/Y$ as required.

But what if we cannot numerate a base, e.g. in the vector space of continuous functions? How could you proceed with infinite dimmensions?

 

[wrobel]

How could you proceed with infinite dimmensions?

who is this addressed to ?

 

[fresh_42]

who is this addressed to ?

The OP of course.

 

[Eclair_de_XII]

vector space of continuous functions

Okay, let $Y=\{\textrm{constant functions}\}$ and let $Z=\{\textrm{functions that depend on }t\}$.

Let $x\in X=C^0(D,R)$ for some domain and range, $D,R$. Then $x(t)=y(t)+z(t)$ for unique functions $y,z$.

$(y_0,\{z_0+y:y\in Y\})\leftarrow x_0$

I'm not really sure what to do, here. Personally, I feel like you could just map $x_0$ to its y-projection in the first coordinate and then associate $x_0$ to the set containing the sum of $x_0$ and some $y$ in the second coordinate. I'm not sure that I'm right, though. My main concern is that there does not seem to be any defined addition or multiplication for the second coordinate of the Cartesian sum.

all vector spaces of the same dimension are isomorphic

You mean because you can map each basis element of one space to exactly one basis element of another, then map it back again?

 

[fresh_42]

You mean because you can map each basis element of one space to exactly one basis element of another, then map it back again?

Yes. If we have bases $\operatorname{span}\{u_k\}\cong\operatorname{span}\{v_k\}$ then we can map $u_k\longmapsto v_k \longmapsto u_k$. In case one of the spaces is $X/Y$ then a basis looks like $x_k+Y$ and we can map $x_k+Y \longmapsto v_k \longmapsto x_k+Y.$

The example with $X=C^0(D,R)$ and $Y=\{f:D\longrightarrow R\,|\,\exists \,r_0\in D \, : \,f(d)=r_0\,\forall\,d\in D\}$ the constant functions (forget $Z$) can be used for the general problem.

Generally we have
\begin{align*}
Y&\stackrel{\iota}{\rightarrowtail} X \stackrel{\pi}{\twoheadrightarrow} X/Y \\
y &\mapsto y \\
\phantom{y}&\phantom{\hookrightarrow\;\;}x\mapsto x+Y
\end{align*}
What we want to find is a injective, linear map $\varphi\, : \,X/Y \longrightarrow X$ such that $\pi\circ\varphi=\operatorname{id}_{X/Y}\,.$

Of course we can simply say $\varphi(x+Y):=x$, but the difficulty is, that $x+Y$ isn't unique. All other $x+y$ with any $y\in Y$ are equivalent to $x,$ i.e. $x+Y=(x+y)+Y.$ So which one shall we use to define $\varphi:$ $\varphi(x+Y)=x$ or $\varphi(x+Y )=x+y\;?$ This means we have to show that we have a well-defined $\varphi .$ Linearity and $\pi\circ\varphi=\operatorname{id}_{X/Y}$ are easy, but why is it well-defined and injective?

 

[Office_Shredder]

Okay, let $Y=\{\textrm{constant functions}\}$ and let $Z=\{\textrm{functions that depend on }t\}$.

Let $x\in X=C^0(D,R)$ for some domain and range, $D,R$. Then $x(t)=y(t)+z(t)$ for unique functions $y,z$.

This is false. For example, 1+t can have y=1, z=t or y=2, z=t-1. You don't have uniqueness.

If you define Z to be functions for which z(0)=0 or something, this might be closer to true.

 

[jbunniii]

Note that $Z = \{\text{functions that depend on }t\}$ is not a subspace. The sum of two non-constant functions (e.g. $t$ and $-t$) can be constant, and multiplying a non-constant function by the scalar zero gives you a constant.

 

[Eclair_de_XII]

why is it well-defined and injective

Well-defined: In the case of $\varphi(x+Y)=x+y$, you are just choosing an element from the set $x+Y$. Whereas in the case of $\varphi(x+Y)=x$, what you're doing is just selecting one of the summands of any given element in $x+Y$.

This is how I see it: let's say we have a 2-D plane. In the latter case, you are choosing an $x$ on the x-axis and not bothering to choose a $y$ on the y-axis. So it's essentially a vertical line, which is not a function, let alone one that is well-defined. Analogously, this vertical line is $x+Y$. In the former case, you are choosing an $x$ and some $y$ on the appropriate axes. This is a point on the 2-D plane and it is a well-defined function.

As for why it is injective... Suppose $\varphi(x_1+Y)=\varphi(x_2+Y)$. Then $x_1+y=x_2+y$. $X$ is closed under the addition, so $x_1=x_2$.

 

[fresh_42]

Well-defined: In the case of $\varphi(x+Y)=x+y$, you are just choosing an element from the set $x+Y$. Whereas in the case of $\varphi(x+Y)=x$, what you're doing is just selecting one of the summands of any given element in $x+Y$.

This is how I see it: let's say we have a 2-D plane. In the latter case, you are choosing an $x$ on the x-axis and not bothering to choose a $y$ on the y-axis. So it's essentially a vertical line, which is not a function, let alone one that is well-defined. Analogously, this vertical line is $x+Y$. In the former case, you are choosing an $x$ and some $y$ on the appropriate axes. This is a point on the 2-D plane and it is a well-defined function.

These ideas use what I was thinking about, too. Either use linear independence: start with $0+Y$, then for $x_1\notin Y$ choose $x_1$, then for $x_2\notin \operatorname{span}\{Y,x_1\}$ choose $x_2$ etc. This comes down to the axiom of choice.

The other geometric view is when we have angles and lengths. Then we can choose orthogonal components as long as there are such.

I can't think of a direct proof at the moment, and I'm not sure there is one, although I suspect it.

As for why it is injective... Suppose $\varphi(x_1+Y)=\varphi(x_2+Y)$. Then $x_1+y=x_2+y$. $X$ is closed under the addition, so $x_1=x_2$.

This is a false argument as you seem to use injectivity to show it. Your "Then" is unclear to me. However, you can use the property $\pi\varphi=1_{X/Y}$. Given $\varphi(x_1+Y)=\varphi(x_2+Y)$ we get $x_1+Y=\pi(\varphi(x_1+Y))=\pi(\varphi(x_2+Y))=x_2+Y$.

We had to show that $\varphi\, : \, X/Y \longrightarrow X$ isinjective, that is: If the images $\varphi(x_i+Y)$ of two elements are equal, then the elements themselves must have been equal, that is $x_1+Y=x_2+Y$. The $x_i$ do not have to be equal. Also you can conclude $a=b \Longrightarrow f(a)=f(b)$ for every well defined function $f$, but not the other way around: $f(a)=f(b) \nrightarrow a=b$. This injectivity which we wanted to prove, so we cannot just cancel $\varphi$ as you did. This would be the conclusion, not the assumption. Applying $\pi$ however is allowed.

 

[wrobel]

I thought that the problem had actually been solved in #3. Strange

 

[fresh_42]

I thought that the problem had actually been solved in #3. Strange

I don't think rephrasing the problem can be considered a solution.

 

[wrobel]

I don't think rephrasing the problem can be considered a solution.

I don't think it was a rephrasing.
I implicitly applied a well-known theorem:) I expected that the prompt would be understood

The well-known theorem is as follows.
Theorem. Every subspace $Y\subset X$ of a vector space $X$ has a compliment space $W\subset X,\quad X=Y\oplus W$.
In the infinite dimensional case ,this follows from the Zorn lemma. I do not think that for infinite dimensional case the problem under discussion can be solved without the Choice axiom.

 

[Eclair_de_XII]

This injectivity which we wanted to prove, so we cannot just cancel $\varphi$ as you did. This would be the conclusion, not the assumption. Applying however is allowed.

What I did was apply the definition of $\varphi(x+Y)$ to the two vectors $x_1,x_2$. You defined $\varphi(x+Y)=x+y$ for some $y$. What I did was apply $\varphi$ to $x_1+Y$ and $x_2+Y$; I got $\varphi(x_1+Y)=x_1+y$ and $\varphi(x_2+Y)=x_2+y$. Then I equated the two expressions on the right, then canceled the $y$ on either side to arrive at the conclusion: $x_1=x_2$. I'm sorry for not being clearer.

 

[fresh_42]

What I did was apply the definition of $\varphi(x+Y)$ to the two vectors $x_1,x_2$. You defined $\varphi(x+Y)=x+y$ for some $y$. What I did was apply $\varphi$ to $x_1+Y$ and $x_2+Y$; I got $\varphi(x_1+Y)=x_1+y$ and $\varphi(x_2+Y)=x_2+y$. Then I equated the two expressions on the right, then canceled the $y$ on either side to arrive at the conclusion: $x_1=x_2$. I'm sorry for not being clearer.

As for why it is injective... Suppose $\varphi(x_1+Y)=\varphi(x_2+Y)$. Then $x_1+y=x_2+y$. $X$ is closed under the addition, so $x_1=x_2$.

No. You only have $x_1-x_2\in Y.$ The images live in $X$, so $Y\neq 0$ here.

If we have $\varphi(x_1+Y)=x_1+y_1=\varphi(x_2)=x_2+y_2$ because we cannot know which $x_i$ represents the coset, then we cannot conclude $x_1=x_2.$ We only know $x_1-x_2\in Y$. You have to use the property $\pi\varphi=1_{X/Y}$ because this is what is required from $\varphi.$ Otherwise any linear function $X/Y\longrightarrow X$ would do, which is not the case.

 

[WWGD]

If we allow the big guns, there is a result that every short exact sequence if vector spaces splits. Think this is what @wrobel was mentioning.

 

[wrobel]

What I mentioned I explained in #15.
Once again: without the Choice axiom, you will not solve the infinite dimensional version of the problem .

 

[fresh_42]

If we allow the big guns, there is a result that every short exact sequence if vector spaces splits. Think this is what @wrobel was mentioning.

Yes, but that is what had to be proven in my opinion.

What I mentioned I explained in #15.
Once again: without the Choice axiom, you will not solve the infinite dimensional version of the problem .

Thanks, that was what I wasn't sure about.

 

[Eclair_de_XII]

I'm confused about what the function $\pi:X\longrightarrow X/Y$ is supposed to do.

I understand that it is a projection. I have normally worked with spaces that consist of finitely many components; the projections I am familiar with usually map elements of such spaces to only one of the components (i.e. mapping a vector in $\mathbb{R}^2$ to either its x- or y-coordinate).

But here, we have a vector $x\in X$ and the space $X/Y$ consists of cosets $x'+Y=\{x'+y:y\in Y\}$. In short, I am confused as to how to decompose the former into the latter.

 

[Eclair_de_XII]

I'm trying to toy around with these ideas, right now, with actual spaces.

I set:

$X=\mathbb{R}^2$
$Y=\{(x,x):x\in \mathbb{R}\}$

$X/Y=\{(x,y)+Y:(x,y)\in X\}$
where
$(x,y)+Y=\{(x,y)+p:p\textrm{ is a point on the line }y = x\}$

Right now, it just looks like $(x,y)+Y$ is an exact copy of $Y$, except shifted by $(x,y)$. Also, for all but one $(x,y)$, $(x,y) + Y$ does not seem to be a subspace of $X$ for the sole reason that it does not have a zero. For example, we can represent $y=x+1$ as:

$(-1,0)+Y$ and $(0,1)+Y$
but
$(-1,0)\neq (0,1)$

which seems to align with what fresh42 said earlier about the non-correlation between $x_1+Y=x_2+Y$ and $x_1=x_2$. Also, the quotient space is like a of collection of the copies of the graph of $1_\mathbb{R}$ shifted by some constant.

If I had to take a guess, $\pi$ is defined something like $\pi:(x,y)+p\mapsto (x,y)\mapsto (x,y)+Y$.

 

[jbunniii]

I'm confused about what the function $\pi:X\longrightarrow X/Y$ is supposed to do.

I understand that it is a projection. I have normally worked with spaces that consist of finitely many components; the projections I am familiar with usually map elements of such spaces to only one of the components (i.e. mapping a vector in $\mathbb{R}^2$ to either its x- or y-coordinate).

But here, we have a vector $x\in X$ and the space $X/Y$ consists of cosets $x'+Y=\{x'+y:y\in Y\}$. In short, I am confused as to how to decompose the former into the latter.

What is the most natural function from $X$ to $X/Y$? (Hint: if $x \in X$, then which coset contains $x$?)

 

[Eclair_de_XII]

Well, $x+Y=\{x+y:y\in Y\}$ contains $x$, since $0\in Y$.

 

[Eclair_de_XII]

Wait, so in the example I described in post #22, would it be like:

$\pi:(x,y)+p\mapsto (x,y)+p+Y$

 

[jbunniii]

Wait, so in the example I described in post #22, would it be like:

$\pi:(x,y)+p\mapsto (x,y)+p+Y$

Doesn't $(x,y)$ refer to an arbitrary point in $X = \mathbb R^2$? If so, then $\pi(x,y)$ would simply be $(x,y) + Y$. In other words, $\pi$ maps each point of $X$ to the coset that contains that point.

Note that if $p \in Y$, then $p+Y = Y$, so $(x,y) + p + Y = (x,y) + Y$.

 

[Eclair_de_XII]

In other words, $\pi$ maps each point of $X$ to the coset that contains that point.

Oh, I think I understand why $\varphi$ is injective. In the case of $X=\mathbb{R}^2$ and $Y=\textrm{graph of }y=x$, $\varphi$ maps two parallel lines (cosets $x_1+Y,x_2+Y$) to the same point (vector $x$). Those lines must be equal, since if these two lines were distinct, they could not contain the same point, since they never intersect.

It's like these cosets are equivalence classes, in the respect that they are either disjoint or identical.