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I am beginning to study algebraic topology. Suppose $X$ and $Y$ are topological spaces. In the lecture notes, the topology of $X \coprod Y$ was defined in the following way: "the open subsets are precisely those of the form $A \coprod B$ with $A \in \tau_X$ and $B \in \tau_Y$".

This tacitly implies that if $A \subseteq X$ and $B \subseteq Y$ then $A \coprod B \subseteq X \coprod Y$. I drew the following diagram for this situation: $\require{AMScd}$ \begin{CD} A @>{i_A}>> A \coprod B @<{i_B}<< B\\ @V{\iota_A}VV @V{\phi}VV @V{\iota_B}VV\\ X @>{i_X}>> X \coprod Y @<{i_Y}<< Y \end{CD} Here $i_A, \iota_A$ etc. are all the normal inclusion maps, and $\phi$ is defined in the natural way, associating $i_B(b)$ with $i_Y(\iota_B(b))$ for all $b \in B$ and $i_A(a)$ with $i_X(\iota_A(a))$ for all $a \in A$. Clearly $\phi$ is an injection, so I can see how $A \coprod B \cong \phi(A \coprod B) \subseteq X \coprod Y$, but I don't see how $A \coprod B \subseteq X \coprod Y$.

In other places on the internet (e.g. Definition of Disjoint Union Spaces) the topology is defined in a slightly different way that seems not to have this issue.

Does the definition in my lecture notes actually make sense? Am I just being too pedantic about pointless details?

Harry Partridge
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    The definition you quote has a parenthetical comment saying "considering this as a subset of the disjoint union". That's exactly what your "issue" is. – Arturo Magidin Jul 28 '21 at 02:29
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    How do you define the disjoint union? Via universal properties, or some specific construction? An easy way to construct the disjoint union of two sets is to define $X\amalg Y$ to be the set $X\times{1}\cup Y\times{2}$, and letting $i_X(x) = (x,1)$, $i_Y(y)=(y,2)$. If you define it this way, then for $A\subseteq X$ and $B\subseteq Y$, you have that $A\amalg B$ is literally a subset of $X\amalg Y$. – Arturo Magidin Jul 28 '21 at 02:31
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    The disjoint union was defined like that, but if X was indexed with 1, Y with 2, then how would we also know to index A with 1 and B with 2? – Harry Partridge Jul 28 '21 at 02:33
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    You are defining the disjoint union of two sets, whatever the two sets may be, as the union of the first set times ${1}$, and the second set times ${2}$. So $A\amalg B$ is the disjoint union of two sets, $A$ the first, $B$ the second, so you would take, as per the definition, the union of $A\times{1}$ and $B\times{2}$. Note, however, that under this literal definition you would have that $B\amalg A$ is not a literal subset of $X\amalg Y$ (in most cases). – Arturo Magidin Jul 28 '21 at 02:35
  • As I said, the quoted definition does make sense to me, but I was wondering if the definition in my lecture notes is also correct – Harry Partridge Jul 28 '21 at 02:36
  • I don't understand why you believe they are different. They look exactly the same to me. – Arturo Magidin Jul 28 '21 at 02:36
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    There are ways of making this construction-independent, and just use the universal property of the disjoint union (which yields a universal property for the topological disjoint union). – Arturo Magidin Jul 28 '21 at 02:39
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    Ok I see now what you are saying with $A \coprod B$ being a literal subset of $X \coprod Y$ when they are constructed in the same order. – Harry Partridge Jul 28 '21 at 02:41
  • The quoted definition is identifying $X$ with its image $i_X(X)$ and $Y$ with its image $i_Y(Y)$. – Arturo Magidin Jul 28 '21 at 02:51
  • I do not think that the tag "algebraic-topology" is adequate here. It is a question about general topology, – Paul Frost Jul 28 '21 at 07:48
  • @ArturoMagidin I am confused: With the definition $X \coprod Y = X \times {1} \cup Y \times {2}$ you state that $A \coprod B$ is not a literal subset of $X \coprod Y$ (second comment) and that $A \coprod B$ is literally a subset of $X \coprod Y$ (first comment). – Paul Frost Jul 28 '21 at 07:58
  • @PaulFrost: the second comment says that $B\amalg A$ (note the order!) is not a literal subset of $X\amalg Y$, not that $A\amalg B$ isn't. – Arturo Magidin Jul 28 '21 at 14:07
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    @ArturoMagidin It seems I was more confused than I thought ;-) – Paul Frost Jul 28 '21 at 23:10

1 Answers1

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The core of your question has not much to do with topology - it is a question about sets. Before you can define a topology on $X \coprod Y$, you have to understand what this object is on the level of sets. Already on that level you can ask (without any reference to topology)

If $A \subset X$ and $B \subset Y$, is it true that $A \coprod B \subset X \coprod Y$?

There are two ways to introduce a new mathematical object:

  1. Direct constructive approach: Simply give an explicit construction of the object.

  2. Conceptual approach: Specify a characteristic property determining the desired object up to unique isomorphism and then prove existence via giving an explicit construction.

In my opinion the conceptual approach should be preferred because it explains the purpose of what is introduced. This approach is of course more abstract - and it also requires to construct the new object.

I think you know that the concept of disjoint union of sets can be introduced as follows:

Given two sets $X,Y$, we say that a triple $(X \coprod Y, i_X, i_Y)$ consisting of a set $X \coprod Y$ and functions $i_X : X \to X \coprod Y, i_Y : Y \to X \coprod Y$ has the disjoint union property if for each pair of functions $f_X : X \to Z, f_Y :Y \to Z$ having the same codomain $Z$ there exists a unique function $f : X \coprod Y \to Z$ such that $f \circ i_X = f_X, f \circ i_Y = f_Y$.

It is easy to verify that all triples having the disjoint union property are isomorphic via a unique "structure preserving" isomorphism. Moreover, the functions $i_X, i_Y$ must necessarily be injective and we have $i_X(X) \cup i_Y(Y) = X \coprod Y, i_X(X) \cap i_Y(Y) = \emptyset$. Therefore they embed $X, Y$ as disjoint subsets into $X \coprod Y$, and frequently authors regard $X,Y$ as genuine subsets of $X \coprod Y$ although this in general not literally true.

It remains to construct a triple with the disjoint union property. A standard approach is to define $$X \coprod Y = X \times \{1\} \cup Y \times \{2\},$$ $$i_X(x) = (x,1), i_Y(y) = (y,2).$$ With this definition the answer to your question is "yes".

Note that if $X \cap Y = \emptyset$ we can also take $X \coprod Y = X \cup Y$; then $i_X, i_Y$ are the genuine inclusions. Then the answer is again "yes". However, we could use a more exotic definition like

$$X \coprod Y = X \times \{(X,1)\} \cup Y \times \{(Y,2)\}$$ with the obvious $i_X, i_Y$. In that case the answer is "no".

You see that the answer to your question depends on the specific construction of disjoint unions. You can do it such that you get genuine inclusions $A \coprod B \subset X \coprod Y$, and this is a particlar nice feature, but can also do it in another way.

In my opinion the property $A \coprod B \subset X \coprod Y$ is not that important. It suffices to know that $A \coprod B$ embeds canonically into $X \coprod Y$, and that is all we need.

Remark:

There are many similar cases, for example the Cartesian product of sets. It also depends on the specific construction whether we have $A \times B \subset X \times Y$.

Update concerning the canonical embedding of $A \coprod B$ into $X \coprod Y$:

Given functions $u : X \to X', v : Y \to Y'$, consider $f'_X = i_{X'} \circ u : X \to X' \coprod Y', f'_Y = i_{Y'} \circ v : Y \to X' \coprod Y'$. By the universal property we get a unique function $u \coprod v : X \coprod Y \to X' \coprod Y'$ such that $(u \coprod v) \circ i_X = f'_X = i_{X'} \circ u, (u \coprod v) \circ i_Y = f'_Y = i_{Y'} \circ v$.

Apply this to the inclusions $j_A : A \hookrightarrow X, j_B : B \hookrightarrow Y$.

Paul Frost
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