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Topic: Smash Product
Replies: 1   Last Post: May 29, 2011 11:30 AM

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 Mike Posts: 96 Registered: 6/2/06
Re: Smash Product
Posted: May 29, 2011 11:30 AM

On Apr 22, 1:30 pm, William Elliot <ma...@rdrop.com> wrote:
> Let X and Y be topological spaces.
>
> The wedge of X and Y at a in X, b in Y is
> .. . (X v Y)_(a,b) = Xx{b} \/ {a}xY
>
> The smash product of X and Y at a in X, b in Y is
> .. . (X ^ Y)_(a,b) = (X x Y)/(X v Y)_(a,b).
>
> where for A subset Z, Z/A is the quotient space of Z
> identifying all the points of A to a single point.
>
> For example.  (R v R)_(0,0) is the union of the x-axis and the y-axis.
>
> What's an intuitive description or understanding of
> (R ^ R)_(0,0) or of ([0,oo) ^ [0,oo))_(0,0) ?

The first smash product is homeomorphic to the wedge of four copies of
X where X is the union of an open disk and one point on its boundary.
The second smash product is homeomorphic to X.

> I use = to mean equal within a homeomorphism.
>
> It's easy to show that the wedge is associative that
> ((X v Y)_(a,b) v Z)_((a,b),c) = (X v (Y v Z)_(b,c))_(a,(b,c))
> = Xx{b}x{c} \/ {a}xYx{c} \/ {a}x{b}xZ
>
> which notates as (X v Y v Z)_(a,b,c).
>
> Is the smash product associative?
> Does ((X ^ Y)_(a,b) ^ Z)_((a,b),c) = (X ^ (Y ^ Z)_(b,c))_(a,(b,c))?

Not if one works with arbitrary based topological spaces. If X = Y is
the discrete set of integers based at 0 and Z is the rationals based
at 0 the triple smash products do not agree. Fact: If X,Y, and Z
are all locally compact Hausdorff then the triple smash products agree
(note that in the above counterexample the rationals is not locally
compact. Generally, for this reason (among others) algebraic
topologists do not work in the category of all topological spaces, but
in a nice category such as compactly generated Hausdorff spaces or
compactly generated weak Hausdorff spaces. In these categories the
smash product is derived from the compactly generated product rather
than the usual cartesian product.

> In addition do both of those smash products
> .. . = (X x Y x Z)/(X v Y v Z)^(a,b,c) ?

Not in general, but yes for compactly generated categories.

> In attempts to proof this, I've discovered and proved a simple theorem
>         X/A / B/A = X/B.
>
> I think for a in A subset X, b in Y, something like
> (X/A x Y)/(X v Y)_(a,b) = (X x Y)/(A v Y)_(a,b)
>
> would suffice.  However something doesn't seem quite right
> about the formulation of the lemma nor does the simple
> theorem seem of use.
>
> Perhaps I need something like if B is a partition of X and
> A s refinement of then X/A / B/A = X/B where now Z/A is the
> quotient space of Z gotten by mapping Z into the partition A.
>
> Then I think the following holds.
> X/A x Y = (X x Y)/R
>
> where R is the partition of X x Y into the equivalence classes of the
> equivalence relation (x,y) ~ (a,b) when y = b & (x = a or x,a in A).
>
> Help, I'm smashed tanking too much of that product.

Date Subject Author
4/22/11 William Elliot
5/29/11 Mike