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

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William Elliot

Posts: 248
Registered: 10/7/08
Smash Product
Posted: Apr 22, 2011 1:30 PM
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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) ?

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))?

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

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
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William Elliot
Read Re: Smash Product

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