1
$\begingroup$

I am trying to understand the concept of wedge sum in the context of Algebraic Topology.

Say I am given two sets, $A=[1, 2, 3]$ and $B=[4, 5]$

Now, disjoint union $A \sqcup B$ = $[(1,0), (2,0), (3,0), (4,1), (5, 1)]$

Now lets identify points $(1,0)$ and $(4,1)$ as similar.

Hence $A\vee B = A \sqcup B $/~ = $[[(1,0), (4,1],(2,0), (3,0), (5, 1)]$

Is the above example correct?

Also a different example, if there are two $S^1$ having two distinct base points, say $x_0$ and $y_0$, and if we identify these two base points as similar, then :

$S^1 \vee S^1 =[($ two base points glued together), $S^1-x_0, S^1-y_0)$

And if this circle example is correct, my question is what happens to the set index, when we construct the disjoint union as depicted in the first example, i.e., elements of first set is appended with $0$, and the elements of second set appended with $1$ as in first example above. Because I have constructed the wedge sum in circle case without explicitly mentioning the set index here. And IF both the example are correct, why the second example does not need the set index.

Also in first example, there is a question of how to read the results. Say the numbers in both the original sets mean length of sticks. Now in the wedge sum each element has two numbers, e.g, $(5,1)$, then what individual element means in the context of length of stick?

$\endgroup$
  • 1
    $\begingroup$ The set index is implied I believe. It's not hard to know which original circle any unglued point came from. $\endgroup$ – Osama Ghani Jul 10 '17 at 7:49
2
$\begingroup$

Both your examples are correct. The index set is usually omitted since it is only a technicality that blurs the vision:

$$S^1\vee S^1=\big(S^1\sqcup S^1\big)/\sim=\big(S^1\times\{0\}\cup S^2\times\{1\}\big)/\sim$$

and formally the relation $\sim$ is induced by (i.e. the smallest equivalence relation for which) $(x_0, 0)\sim (y_0, 1)$

Also in first example, there is a question of how to read the results.

Well, you should not think about the second coordinate, since (as I've said) it's only a technicality. On the other hand the meaning of a wedge sum is a whole different story. It's hard to say without bigger context.

$\endgroup$
0
$\begingroup$

One way to think about the wedge sum of two spaces is as a subspace of their product

$X\vee Y\cong \{(x,\ast_X),(\ast_Y, y))\mid x\in X, y\in Y\}=(X\times\ast_Y)\cup (\ast_X\times Y)\subset X\times Y,$

where $\ast_X$, $\ast_Y$ are the respective basepoints of $X$, $Y$. This can be be generalised to wedge sums indexed over finite sets. That is,

$\bigvee_{i\in \mathcal{I}} X_i\cong\{(\ast,\dots\ast,x_i,\ast,\dots,\ast)\mid x_i\in X_i,\; \text{some}\; i\in\mathcal{I}\}\subset \prod_{i\in\mathcal{I}}X_i$.

Then the index just tells you which coordinate in the product is not necessarily a basepoint. Note that if the $i$-indexed point $x_i=\ast$ then it is identified with the $j$-indexed basepoint $x_j=\ast$, so this is what we want. Feel free to review the construction of the cartesian product of sets, to see how your intuition was correct for your first example.

Note that here may be some difficulty with larger index sets due to the way in which the various spaces are topologised, but when your index set is finite there are no problems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.