# Gracious Living

Product and Disjoint Union Topologies
November 8, 2010, 15:55
Filed under: Math, Topology | Tags: , , , ,

As the subspace topology is the “best way” to topologize a subset, the product and disjoint union topologies are the “natural ways” to topologize Cartesian products and disjoint unions of topological spaces.  Usually the disjoint union topology is only glossed over, while more time is spent on the product topology; I’m introducing them together in order to show you some of the similarities between them addressed by category theory.  The relationship is one of duality, something like the intersection and union of sets.  There’s this goofy mathematician way of putting “co” in front of dual constructions, so we could also call the disjoint union the “coproduct”.

First, remember the corresponding ideas for sets: a Cartesian product of sets $\prod_{\alpha\in A}X_\alpha$ (tiny version $X\times Y$)is the set of ordered tuples $(x_\alpha)_{\alpha\in A}$ (resp. $(x,y)$).  A disjoint union of sets $\coprod_{\alpha\in A}X_\alpha$ (resp. $x\sqcup y$) is just their union, assuming a priori that they’re disjoint.  One way of forcing this is by treating it as $\bigcup_{\alpha\in A}\left(X_\alpha\times\{\alpha\}\right)\subset\bigcup_{\alpha\in A}\left(X_\alpha\times A\right)$.  So $\mathbb{R}\sqcup\mathbb{Q}$ isn’t $\mathbb{R}\cup\mathbb{Q}=\mathbb{R}$; it actually looks more like $\mathbb{R}\times\{0\}\cup\mathbb{Q}\times\{1\}\subset\mathbb{R}\times\{0,1\}\cup\mathbb{Q}\times\{0,1\}$.  We have all the reals on one level, and all the rationals again on the level above that.

From a categorical perspective, the important thing in both constructions is the set of maps you get out of them.  For a product, there are canonical projections, again indexed by $\alpha\in A$, of the form $p_\alpha:\left(\prod X_\alpha\right)\rightarrow X_\alpha$, where $p_\alpha$ sends every tuple to its $\alpha$th coordinate.  For a disjoint union, there are canonical injections of the form $i_\alpha:X_\alpha\rightarrow\left(\coprod X_\alpha\right)$ that send $x\in X_\alpha$ to $(x,\alpha)\in\coprod X_\alpha$.

An example of how I think of it is shown above.  The blue things are our starting sets, and the red things are the product and disjoint union.  The black arrows are canonical injections into the disjoint union and projections from the product.  The green arrows are examples of an injection into the product and a projection from the disjoint union, but they’re highly non-canonical.

Okay.  Remember from the continuity post that a continuous map $f:X\rightarrow Y$ stays continuous if $X$ takes a finer topology, or $Y$ a coarser one.  Given topological spaces $X_\alpha$, if we’re to topologize their product or disjoint union, we want to make the projections or injections continuous.  The canonical way to do this, then, is: give the product the coarsest topology such that all the projections are continuous; give the disjoint union the finest topology such that all the injections are continuous.

This is a lovely general way of putting it, but fortunately we can say more concrete things about their topologies.  Let’s start with the disjoint union of $X_\alpha$.  We can identify each $X_\alpha$ with its “piece” of the disjoint union via its canonical injection.  Probably what we should do is just give the “piece” the same open sets as the original thing.  In fact, this makes each injection continuous: an open set in $i_\alpha(X_\alpha)$ has preimage, well, itself in $X_\alpha$; and if $\beta\ne\alpha$, then an open set in $i_\beta(X_\beta)$ has preimage $\emptyset$ under $i_\alpha$, which is open!  And if we added another open set to the disjoint union, then it must have a nonempty intersection with some $i_\alpha(X_\alpha)$; then $i_\alpha$ is no longer continuous, since that set pulls back to something non-open in $X_\alpha$.  So we can summarize the disjoint union topology as the union of the topologies of all the $X_\alpha$, each applied to $i_\alpha(X_\alpha)$.  This means that the subspace topology of each $i_\alpha(X_\alpha)$ agrees with the topology on $X_\alpha$; but this should make sense, because we had a similar way of expressing the subspace topology on $X_\alpha$!

The product topology is a tad trickier.  Let’s start with two spaces, $X$ and $Y$.  The preimage of an open set $U\subset X$ is $U\times Y$, and that of $V\subset Y$ is $X\times V$.  So we want these to be open to make the projections continuous.  Then their intersection, $U\times V$, must be open as well.  Sets of the form $U\times V$ thus form a basis for the product topology.  If we took away an open set from this, we’d have to take away a set of the form $U\times Y$ or $X\times V$, and then the corresponding projection would be discontinuous.  So this is the coarsest topology making all these guys continuous.

The trouble is when we get to infinite products.  The subbasic open sets of $\prod X_\alpha$ are of the form $\dotsb\times X_\beta\times U\times X_\gamma\times\dotsb$.  That is, we replace one $X_\alpha$ in the giant product with one of its open subsets.  But recall that open sets are only closed under finite intersections.  Thus, a basis for the product topology here is given by $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for all $\alpha$, and where for all but finitely many $\alpha$, $U_\alpha=X_\alpha$.

To assuage your fears, let me point out that if we remove the cofiniteness restriction and allow open sets of the form $U\times V\times W\times\dotsb$, we still get a topology.  It’s called the box topology.  Obviously, for infinite products, it’s strictly finer than the product topology, since any basis element of the product topology is also a basis element of the box topology, but not vice versa.  I’d argue that it’s less useful, in the end, too.  An example from Munkres that I feel is worth repeating: let $\mathbb{R}^{\omega}$ be the product of countably many copies of $\mathbb{R}$, and let $f(x)=(x,x,x,\dotsc)$.  If we compose $f$ with any projection map, we clearly get a continuous map.  Because of this, $f$ itself must be continuous under the product topology (check this by looking at the preimages of subbasis elements).  But under the box topology, the preimage of, say, $(1,-1)\times(1/2,-1/2)\times(1/3,-1/3)\times\dotsc$ is just $\{0\}$, which is not open!  Thus, $f$ fails to be continuous under the box topology.

One use of the product topology is studying function spaces, which are topological spaces whose points are actually functions.  We’ll go into more detail about this later, but for now I’m just defending the “exponent” notation for sets of functions.  The space $\mathbb{R}^{[0,1]}$ is the set of functions from $[0,1]$ to $\mathbb{R}$.  We can also look at it as a cartesian product of $\mathbb{R}$ indexed by $[0,1]$; an element is then a tuple, which is an assignment of an element of $\mathbb{R}$ to every element of $[0,1]$, which is just a function!  But we can also give $\mathbb{R}^{[0,1]}$ the product topology, and it turns out that a set of “points” (really functions) converges if and only if the functions converge pointwise (i. e. for a given $x\in [0,1]$, the sequence $f_i(x)$ converges).  Looking at the elements of $\mathbb{R}^{[0,1]}$ as tuples again, it’s clear that this is just a corollary of the fact that a sequence in the product space converges iff all its projections converge — which is in turn a corollary of the fact you proved above, that a function to the product space is continuous iff its composition with every projection function is continuous!

To wrap us up, just some obvious facts, which you should check if you feel like: a basis for each $X_\alpha$ gives a basis for their disjoint union, product, or box topology, in the way you’d expect, and likewise for subbases.  The sets $A\times(B\times C)$ and $(A\times B)\times C$ are not equal (since $(a,(b,c))\ne ((a,b),c)$), but they are canonically homeomorphic (via the map $(a,(b,c))\mapsto ((a,b),c)$).  Ditto for the disjoint union if it’s defined “formally” (as a subset of the product with $\{0,1\}$ or whatever).  A product of subspaces is a subspace of the product, and a subspace of a set is a subspace of a disjoint union containing that set.  Oh, and each $X_\alpha$ is open in their disjoint union, which is useful to remember.  By the way, $\mathbb{N}$ has the discrete topology as a subspace of $\mathbb{R}$, so it’s the disjoint union of its points.  $\mathbb{Z}$ is the same, but $\mathbb{Q}$ is not.

I think I’ll talk about connectedness tomorrow, and hold off on quotients for a little bit.