Suppose $Z$ is a linear space and $Y \subseteq Z$ is a linear subspace. Then we can define the quotient space $Z\mid_Y$, i.e. $Z$ modulo $Y$, as follows:

$Z\mid_Y = \{ [z]_Y \mid z \in Z \}$

Where $[z]_Y$ is the equivalence class of $z$ for the relation $\sim^Y$ defined on $Z$ by:

$x \sim^Y y \equiv x - y \in Y$

This agrees with the Wikipedia definition of quotient space.

Question: Suppose $X_0 \subseteq Z$ is another linear subspace of $Z$. Then how do we define the quotient space $X_0\mid_Y$? We can't just apply the previous definition because, in general, we do not have that $Y \subseteq X_0$. However, I believe we can just relax that restriction and do this:

$X_0\mid_Y = \{ [x]_Y \mid x \in X_0 \}$

I'm 99% percent sure that this definition is the correct one, but hoping someone can point me to a definition of quotient space that doesn't require $Y$ to be a subset of the space being "divided"?

What's slightly odd to me is that the equivalence classes in this resulting quotient space are not necessarily subsets of $X_0$, so it goes against the intuition that $Y$ is somehow "dividing" up $X_0$.

Alternative Formulation: Cosets

As per this question, we can also define the quotient space $Z\mid_Y$ as follows:

$Z\mid_Y = \{ z + Y \mid z \in Z \}$

Where $z + Y = \{z + y \mid y \in Y\}$ is the coset of $Y$ containing $Z$. In this case, what I'm asking is can we define the quotient space for $X_0$ in general, without necessarily having $Y \subseteq X_0$, as follows:

$X_0\mid_Y = \{ x + Y | x \in X_0 \}$

This definition seems natural to me - the only unnatural part is the term "quotient" in the term "quotient space", and the division-like notation in $X_0\mid_Y$, because we're not necessarily dividing up $X_0$, rather we're sort of mapping it over to a set of cosets. But I suppose since the term "quotient" makes sense for the case when $Y \subseteq Z$, maybe we just overload it to apply also to this case? Or is this called something else?


I came across use of this version of quotient at the end of this video, where the professor equates the completion of a space to the closure of the "quotient" of its embedded space:

VIDEO: Completing a normed linear space. Time: 1:01:33.

EDIT: Reason for my Confusion

I initially thought that:

$X_0\mid_Y = X_0\mid_{Y \cap X_0}$

The reasoning being:

  1. $Y$ and $X_0$ are linear spaces
  2. The intersection of linear spaces always yields a linear space
  3. $Y \cap X_0 \subseteq X_0$, so we can apply the original definition

However, in general, this is not the case. That is, there exists some example such that:

$X_0\mid_Y \neq X_0\mid_{Y \cap X_0}$

In fact, the example in the linked lecture video is such an example:

$X_0$ is the set of constant sequences of the form $(x, x, x, \dots)$. $Y$ is the set of sequences $(y_1, y_2, y_3, \dots)$ which are equivalent to $(0, 0, 0, \dots)$ in the sense that limit of the norms of elements $y_i$ converge to zero. So in that case $Y \cap X_0$ is the singleton set:

$Y \cap X_0 = \{(0, 0, 0, \dots)\}$

Which certainly is not equal to $Y$, as long as there is some non-zero element $x \in X$, because we can define $s = (x, 0, 0, \dots)$ and we know that $s \in Y$ but clearly $s \notin Y \cap X_0$ because $s \neq (0, 0, 0, \dots)$.

So then, for example, to show that

$X_0\mid_Y \neq X_0\mid_{Y \cap X_0}$

We just have to show that the equivalence classes differ for one element. Choose $0$ to be that element:

$[0]_Y = Y \neq Y \cap X_0 = [0]_{Y \cap X_0}$

So the quotient spaces are not equal.


1 Answer 1


Once you chose your linear space $Z$ and your subspace $Y\subset Z$, you have a linear (and onto) map $\pi:Z\to Z|_Y$ given by $z\mapsto [z]_Y$. Since a linear map sends linear subspaces to linear subspaces, you can see that both of your definitions agree with $X_0|Y=\pi(X_0)$. An equivalent way to define this would be $X_0|_Y:=X_0|_{Y\cap X_0}$, since $X_0\cap Y$ is then a linear subspace of $X_0$.

  • 1
    $\begingroup$ Thanks. I think there's a problem though with this though: "An equivalent way to define this would be ". It was what was confusing me in the first place. I've updated my question accordingly. $\endgroup$ Dec 7, 2021 at 16:26
  • $\begingroup$ The quotient spaces are not equal in the sense that they have the same elements, but rather in the sense that they are isomorphic: the linear map $L:X_0|_{Y\cap X_0}\to \pi(X_0)$ given by $[x]_{Y\cap X_0}\mapsto[x]_Y$ is onto by definition, and injective since $L([x]_{Y\cap X_0})=0$ implies that $[x]_Y=0$, then $x\in Y$, so $x\in X_0\cap Y$ and thus $[x]_{Y\cap X_0}=0$. $\endgroup$
    – Balloon
    Dec 7, 2021 at 17:09
  • $\begingroup$ Thanks. The isomorphism makes sense. I guess though the heart of the question is this. Strictly speaking, do we define: $X_0|_Y:=X_0|_{Y\cap X_0}$. Or do we define it as the image under $L$: $X_0|_Y:=L(X_0|_{Y\cap X_0}) = \pi(X_0)$. I'm assuming the latter, based on the lecturer's comments. Otherwise his proof doesn't make sense. But I can't find an official definition anywhere. I guess the question is more focused on is there a reference where this object is officially defined - the quotient of a space modulo some non-subspace, both in the context of a larger space that contains both. $\endgroup$ Dec 7, 2021 at 18:08
  • $\begingroup$ Yes, I agree! I want to add that since the isomorphism is canonical, it can be made precise to think about these spaces as the same spaces in some sense (in a categorical way). Also think about how we identify $\mathbb{R}$ and $\mathbb{R}^1$ where the latter is rigorously defined as the set of functions $f:\{1\}\to\mathbb{R}$ endowed with natural operations, I think these are problems of same nature. $\endgroup$
    – Balloon
    Dec 7, 2021 at 18:17
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    $\begingroup$ Cool. I think I see your answer in a slightly different light now. I think what you're saying is "While the thing defined as $X_0 \mid_Y$ is not officially a quotient space, as per the Wiki definition, $X_0 \mid_{Y \cap X_0}$ is, and moreover, it's isomorphic to the first, so they're practically the same thing for all intents and purposes." $\endgroup$ Dec 8, 2021 at 8:24

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