0
$\begingroup$

Firs let me to recall the definition of wedge sum:

Let ${\{(X_i,x_i)\}}_{i\in I}$ be a family of pointed topological spaces, i.e., topological spaces with distinguished base points $x_i$. The wedge sum of the family is the quotient space of the disjoint union of members of the family by the identification all $x_i$, i.e, $$\bigvee _{i}X_{i}=\coprod _{i}X_{i}\;/{\{x_i: i\in I\}}$$. In other words, the wedge sum is the joining of several spaces at a single point.

As we know, the wedge sum is sensitive to the choice of the base points. But, I saw an exercise in an algebraic topology book as follows:

Ex. Prove that if $X$ and $Y$ are path-connected topological space, the the wedge some is independent of choosing base points.

I have doubt that this exercise is true!! For example, if we consider a cone $\mathcal{C}$ with (1,0) appex over the set $A=\{(\frac{1}{n},0): n\in\mathbb{N}\}\cup\{(o,o)\}$, then $(X, (0,0))\vee (X, (0,0))$ is not contractible, while $(X, (1,0))\vee (X, (1,0))$ is contractible!!

Now my question is that:

When wedge sum of path-connected topological spaces are independent of choosing base points up to homotopy? For instance, is the mentioned Ex true for the case of path-connected simplicial complexes?

$\endgroup$
  • $\begingroup$ What's the path from (0,0) to (1,0)? $\endgroup$ – Randall Aug 19 '17 at 12:44
  • $\begingroup$ (0,0) is connected to (1,0) by the vertical segment, i.e, $\{(0,y) : 0\leq y\leq 1\}$ $\endgroup$ – MathFun Aug 19 '17 at 17:23
  • $\begingroup$ I understand better now. I now have the same question as you and would speculate that your source is just wrong. Or did they make a blanket assumption earlier that all spaces are well pointed? $\endgroup$ – Randall Aug 19 '17 at 21:02
1
$\begingroup$

I suppose if you only consider well-pointed spaces the wedge sum would be independent of the base point, up to homotopy.

The wedge sum of $(X,x)$ with $(Y,y)$ is the (ordinary) pushout of a diagram $X \xleftarrow{x} * \xrightarrow{y} Y$. Note that a path between two points, say $x, x' \in X$ is precisely a homotopy between $x, x': * \times I \to X$. To get something that is homotopically meaningful, i.e., something that is invariant under replacing the maps in the pushout diagram by homotopic ones, one should really consider the homotopy pushout instead of the ordinary pushout.

In the case $(X,x)$ and $(Y,y)$ are well-pointed, i.e., $\{x\} \hookrightarrow X$ and $\{y\} \hookrightarrow Y$ are cofibrations, the homotopy pushout agrees with the ordinary pushout. This explains why if all the spaces involved are well-pointed, the wedge sum should be independent of the choice of basepoints.

For example, the inclusion of a vertex into a CW complex gives a well-pointed space, almost by definition.

In case you were wondering, the homotopy pushout of $X \xleftarrow{x} * \xrightarrow{y} Y$ is the space formed by taking disjoint union of $X$ and $Y$ and then joining the points $x$ and $y$ with a path. Indeed, a cofibrant replacement of a map $* \xrightarrow{x} X$ can be obtained by attaching a "whisker" at $x$.

$\endgroup$

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.