Confusion about quotient spaces

[AxiomOfChoice]

If $X$ is a topological vector space and $Y$ is a subspace, we can define the quotient space $X/Y$ as the set of all cosets $x + Y$ of elements of $X$. There is an associated mapping $\pi$, called the quotient map, defined by $\pi(x) = x + Y$. If I'm not mistaken, there is an equivalence relation lurking here, too: $x \sim y$ iff $\pi (x) = \pi(y)$.

Here's my question: We know that if $f$ is some function, then $x\in f^{-1}(A)$ if and only if $f(x) \in A$. This is fine - the object on the left of the $\in$ is a point, and the object on the right is a set. But if one tries to apply this to the quotient map and a subset $V\subset X$, we have $x \in \pi^{-1}(\pi(V))$ iff $\pi(x) \in \pi(V)$. The object on the left of the $\in$ here is a set; the object on the right is a set. So what the heck is this supposed to mean? Did the $\in$ turn into a $\subset$ somehow?

 

[Robert1986]

If $X$ is a topological vector space and $Y$ is a subspace, we can define the quotient space $X/Y$ as the set of all cosets $x + Y$ of elements of $X$. There is an associated mapping $\pi$, called the quotient map, defined by $\pi(x) = x + Y$. If I'm not mistaken, there is an equivalence relation lurking here, too: $x \sim y$ iff $\pi (x) = \pi(y)$.

Here's my question: We know that if $f$ is some function, then $x\in f^{-1}(A)$ if and only if $f(x) \in A$. This is fine - the object on the left of the $\in$ is a point, and the object on the right is a set. But if one tries to apply this to the quotient map and a subset $V\subset X$, we have $x \in \pi^{-1}(\pi(V))$ iff $\pi(x) \in \pi(V)$. The object on the left of the $\in$ here is a set; the object on the right is a set. So what the heck is this supposed to mean? Did the $\in$ turn into a $\subset$ somehow?

Have you ever seen Inception?? What I am about to explain is kind of like inception.

The set $\pi(V)$ is actually a set whose elements are sets, themselves. Thus, the coset $x+Y$ is actually an element of the quotient space $X/Y$. HOWEVER, you must be careful when talking about the inverse of this projection. The function $\pi$ is NOT injective. For example, say $V = \{x,y\}$ then $\pi(V)=\{x+Y,y+Y\}$. Now, you might be tempted to say that $\pi^{-1} \pi(V) = \{x,y\}$ but this need not be.

 

[Fredrik]

Since $\pi:X\rightarrow X/Y$ and $V\subset X$, $\pi(x)\in\pi(V)$ indicates that a member of X/Y is a member of a subset of X/Y. The fact that the members of X/Y are subsets of X is irrelevant.

 

[Robert1986]

The fact that the members of X/Y are subsets of X is irrelevant.

In what sense?

 

[Fredrik]

In what sense?

In every sense, but specifically in the sense that it's not a reason to think that the formula $\pi(x)\in\pi(V)$ indicates that there's something really weird going on. (Post #1 suggests that he expects to never see a formula $A\in B$ where A is a set).

 

[Robert1986]

OK. I agree that it is irrelevant when it comes to the OP's question (in the sense that nothing weird is going on). However, the fact that the elements of the quotient space are sets is not irrelevant in every sense, is it? For example, when defining operations in a quotient space one must be careful to make sure that the operations are well defined. Same thing when describing maps from the quotient space to somewhere else. So, while I agree that it might be irrelevant, in some sense, w.r.t the OP's question, I can't agree that it is irrelevant in every sense.

 

[AxiomOfChoice]

Ok, thanks guys. I've been going through the section on quotient spaces in Rudin's Functional Analysis and have managed to confuse myself in a few spots regarding this issue. For example...suppose I want to show $\pi^{-1}(\pi(x)) = Y + x$. I might try to do this by first showing $\pi^{-1}(\pi(x)) \subset Y + x$. So if $z\in \pi^{-1}(\pi(x))$, then $\pi(z) \in \pi(x)$, right? So does that mean that $z + Y \subset x + Y$, both considered to be subsets of $X$? And wouldn't that imply the existence of $y_1,y_2\in Y$ such that $x+y_1 = z+y_2$; i.e., $x + Y \ni x + (y_1 - y_2) = z$? Is there an easier/more elegant way to show this?

 

[AxiomOfChoice]

And the reverse inclusion would go something like so: $z\in Y + x \Rightarrow \pi(z) \in \pi(Y) + \pi(x) = Y + \pi(x)$, so $z + Y \subset Y + x + Y \subset x + Y$; hence $\pi(z) \in \pi(x)$; hence $z \in \pi^{-1}(\pi(x))$. Does that look right?

 

[AxiomOfChoice]

Also, is it accurate to say that if $V\subset X$, then $\pi(V) = \bigcup\limits_{v\in V} v + Y = \bigcup\limits_{y\in Y} y + V$?

 

[Fredrik]

Now I see what's confusing you, because it confused me too when I started writing this reply. We have $\pi(x)=x+Y$, by definition of $\pi$. So the equality you wrote as $\pi^{-1}(\pi(x))=x+Y$ can also be written as $\pi^{-1}(x+Y)=x+Y$. Since x+Y is an element of the codomain of $\pi$, not a subset, we have $\pi^{-1}(x+Y)=\{y\in X|\pi(y)=x+Y\}$. Note the equality sign where you wrote a $\in$ symbol. So the proof of the first identity you mentioned is $$z\in\pi^{-1}(y+X)\ \Leftrightarrow\ \pi(z)=x+Y\ \Leftrightarrow\ z+Y= x+Y\ \Leftrightarrow\ z\in y+Y.$$

Also, is it accurate to say that if $V\subset X$, then $\pi(V) = \bigcup\limits_{v\in V} v + Y = \bigcup\limits_{y\in Y} y + V$?

No. $\pi(V)$ is a subset of X/Y, and $\bigcup_{v\in V}(v+Y)$ is a subset of X.