In the topology book I'm reading I found the following statement:

Let $(X, x_0), (Y, y_0)\in Top_*.$ The "smash product" of $(X, x_0), (Y, y_0)$ is defined as

$$X\wedge Y := X\times Y/X\vee Y,$$ where $$X\vee Y:= X \bigsqcup Y/\sim = X\bigsqcup Y/\{x_0, y_0\}$$

My question is why $$X\wedge Y = (X\times Y)/(X\times \{y_0\})\cup (\{x_0\}\times Y)$$

I don't know how to prove it and I hope, that someone can help.


Your definition of $X\wedge Y$ is an abuse of notation (albeit a common one), and the actual definition is that \begin{equation} X\wedge Y = (X\times Y)/(X\times \{y_0\})\cup (\{x_0\}\times Y).\tag{1} \end{equation} Indeed, the "definition" \begin{equation} X\wedge Y := X\times Y/X\vee Y \tag{2} \end{equation} is strictly speaking meaningless since $X\vee Y$ is not a subset of $X\times Y$. Statement (2) is actually just an informal shorthand for statement (1), since $X\vee Y$ is homeomorphic to the subset $(X\times \{y_0\})\cup (\{x_0\}\times Y)\subseteq X\times Y$ (the map $f:X\coprod Y\to(X\times \{y_0\})\cup (\{x_0\}\times Y)$ given by $f(x)=(x,y_0)$ for $x\in X$ and $f(y)=(x_0,y)$ for $y\in Y$ satisfies $f(x_0)=f(y_0)$ and hence induces a map $g:X\vee Y\to (X\times \{y_0\})\cup (\{x_0\}\times Y)$, which you can show is a homeomorphism). So when your book writes (2), what it actually literally means is (1).

  • $\begingroup$ @ Eric Wofsey. Why $X\vee Y$ is homeomorphic to the subset $X\times \{y_0\} \cup \{x_0\} \times Y$ Can you explain for me? Thank you very much! $\endgroup$
    – longhoang
    Nov 30 '16 at 1:28
  • $\begingroup$ I have added a construction of the homeomorphism. $\endgroup$ Nov 30 '16 at 2:05
  • $\begingroup$ @ Eric Wofsey. Because $X \bigsqcup Y = (X \times \{X\})\bigcup (Y \times \{Y\}) = \{(x, \{X\}\,\,|\,\,x\in X\}\bigcup \{(y, \{Y\}\,\,|\,\,y\in Y\}$ so the map $f: X \bigsqcup Y\to (X\times \{y_0\})\cup (\{x_0\}\times Y)$ given by $f(x; \{X\}) = (x; y_0)$ for $x\in X$ and $f(y; \{Y\}) = (x_0; y)$ for $y\in Y$ satisfies $f(x_0; \{X\}) = f(y_0;\{Y\}).$ right or wrong?? $\endgroup$
    – longhoang
    Nov 30 '16 at 2:32
  • $\begingroup$ Sure, if that's how you define $X\bigsqcup Y$. $\endgroup$ Nov 30 '16 at 2:45
  • $\begingroup$ @ Eric Wofsey. I have a question at math.stackexchange.com/questions/2035807/… Can you prove that my problem? Thank you very much! $\endgroup$
    – longhoang
    Nov 30 '16 at 2:55

Although the question is old now, let me remark that the wedge $(X,x_0) \vee (Y,y_0)$ is the sum, or coproduct, of $(X, x_0), (Y, y_0)$ in the category $Top_*$. Recall that a categorical sum of objects $A_1,A_2$ consists of an object $A_1 + A_2$ and two morphisms $i_k : A_k \to A_1 + A_2$ with a certain universal property which determines the "sum triple" $(A_1 + A_2,i_1,i_2)$ uniquely up to isomorphism.

There are various concrete constructions of the wedge $\vee$. Let us denote by $\vee_{prod}$ the following:

$$(X,x_0) \vee_{prod} (Y,y_0) = X\times \{y_0\}\cup \{x_0\}\times Y$$ with the subspace topology from $X \times Y$ and basepoint $(x_0,y_0)$. The map $i_X : (X,x_0) \to (X,x_0) \vee_{prod} (Y,y_0)$ is given by $i_X(x) = (x,y_0)$, similarly $i_Y$.

We have a map

$$j_X : (X,x_0) \to (X,x_0) \times(Y,y_0) = (X \times Y, (x_0,y_0)), j_X(x) = (x,y_0) ,$$

similarly $j_Y$. Given any wedge construction $\vee$, the universal property of the sum produces a unique map $j : (X,x_0) \vee (Y,y_0) \to (X,x_0) \times(Y,y_0)$ such that $j \circ i_X = j_X, j \circ i_Y = j_Y$.

The smash product is then defined as the pushout of the pair of maps $(j,c)$, where $c : (X,x_0) \vee (Y,y_0) \to \ast$ = one-point space. The map at the opposite side of $c$ is a quotient map $\hat{c} : (X,x_0) \times(Y,y_0) \to (X,x_0) \wedge (Y,y_0)$.

If you look at the standard construction of the pushout as a quotient of the product $(X,x_0) \times(Y,y_0) \times \ast$ and take $\vee_{prod}$ as wedge construction, you will see it results precisely in your concrete definition which we denote for the moment by $(X,x_0) \wedge_{prod} (Y,y_0)$.

You can alternatively take any other wedge $\vee$ and prove that the quotient map $p : (X,x_0) \times(Y,y_0) \to (X,x_0) \wedge_{prod} (Y,y_0)$ and the unique pointed map $a : \ast \to (X,x_0) \wedge_{prod} (Y,y_0)$ complete $(j,c)$ to a pushout diagram. This is done by verifying the universal property of the pushout using the universal property of the sum.


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.