# Defining the Smash product

Given a topological space $X$, in the definition I'm using (from John Lee's Introduction to Topological Manifolds), the quotient space $X/A$ formed by collapsing $A$ to a point, requires $A$ to be a subset of $X$.

The following passage is taken from Algebraic Topology by Allen Hatcher Now there seems to be a few technicalities in the definition that I can't quite wrap my head around.

Given two topological spaces $X$ and $Y$, and base points $x_0 \in X$ and $y_0 \in Y$ we first define the disjoint union to be $X \sqcup Y = \{(\alpha, 1) \ | \alpha \in X \} \cup \{(\alpha, 1) \ | \alpha \in Y \}$ and then we define the wedge product to be $$X \vee Y = X \sqcup Y / \{(x_0, 1), (y_0, 2)\} = \{ [(x_0, 1)]\} \cup \left\{[x'] \ | \ x' \in X \sqcup Y\right\}$$

Then $X \vee Y$, set-theoretically is a partition of $X \sqcup Y$, and this brings about a big set-theoretic issue. Based on my definition right at the start of this page $X \times Y / X \vee Y$ is not defined because $X \vee Y$ is not a subset of $X \times Y$.

It could turn out that $X \vee Y$ is bijective to some subset of $X \times Y$ (and I'm sure this is the case), so if $f : X \vee Y \to X \times Y$ is some bijection, then are the authors forming the smash product as the following? $$X \times Y / f\left[X \vee Y\right]$$

• As stated in the passage from Hatcher, the isomorphism is between $X\vee Y = (X\sqcup Y)/\sim$ (where $\sim$ is defined by $x_0\sim y_0$) and $(X, y_0)\cup (x_0, Y)\subset X\times Y$. – Michael Lee Jun 26 '18 at 11:46
• Also, please forgive my shorthand. $(X, y_0) := X\times \{y_0\}$, $(x_0, Y) := \{x_0\}\times Y$. – Michael Lee Jun 26 '18 at 11:57
• @MichaelLee Isomorphism in which category, $\textbf{Set}$ or $\textbf{Top}_{*}$? – Perturbative Jun 26 '18 at 18:57

Hatcher already explained how $X\vee Y$ (defined classically as the disjoint union with two points collapsed) is a subset of $X\times Y$. You even quoted him yourself!

To be more precise: let $x_0\in X$ and $y_0\in Y$ be fixed points and consider

$$f:X\sqcup Y\to X\times\{y_0\}\cup\{x_0\}\times Y$$ $$f(x,1)=(x,y_0)$$ $$f(y,2)=(x_0, y)$$

You can easily verify that this is a quotient map. So now if $v,w\in X\sqcup Y$ then we have $v\sim w$ if and only if $f(v)=f(w)$. You can easily see that $\sim$ coincides with $\{(x_0,1), (y_0,2)\}$ collapse and so

$$(X\sqcup Y)/\sim=(X\sqcup Y)/\{(x_0,1), (y_0,2)\}=X\vee Y$$

But thanks to general topology we know that quotient maps induce homeomorphisms from $\sim$ relation:

$$F:X\vee Y \to X\times\{y_0\}\cup\{x_0\}\times Y$$ $$F([v]_\sim)=f(v)$$

In that way $X\vee Y$ becomes $X\times\{y_0\}\cup\{x_0\}\times Y$.

It could turn out that $X \vee Y$ is bijective to some subset of $X \times Y$ (and I'm sure this is the case), so if $f : X \vee Y \to X \times Y$ is some bijection, then are the authors forming the smash product as the following? $$X \times Y / f\left[X \vee Y\right]$$

Yes and no. For Hatcher $X\vee Y:=X\times\{y_0\}\cup\{x_0\}\times Y$ is simply a definition, so indeed it is a subset of $X\times Y$ to begin with. But if you start from the disjoint union definition then yes, with $f:=F$ as defined by me earlier.

• Ahh so $X \vee Y$ is homromorphic to $X \times \{y_o\} \cup \{x_0\} \times Y$ and we then collapse $X \times \{y_o\} \cup \{x_0\} \times Y$? – Perturbative Jun 26 '18 at 19:04

what would be a more precise way to write is the following: We consinder $(X,x_0),(Y,y_0)$ as based spaces, i.e. we fix for both spaces one specific point. Then $X \wedge Y=X\times Y/ \sim$ where $\sim$ identifies all points of the subspaces $X\times y_0$ and $x_0 \times Y$ which each other, so we collapse the subset $X\times y_0 \sqcup x_0 \times Y$ that can be seen as $X \lor Y$, as it is a copy of $X$ and $Y$ only intersecting in one point $(x_0,y_0)$. As an example, the 2.dim. torus $S^1 \times S^1$ gets smashed to the sphere $S^2$ where we collapse two spheres joint at one chosen point sitting in the torus.

It is easy to show that both definitions are binary coproducts in the category $\mathsf{Top}_\ast$. Thus they are homeomorphic.

• It seems to me that showing that both are coproducts is not harder or easier than the current problem. – Andres Mejia Jun 26 '18 at 15:30