Smash products distributes over wedge

Question

In Bert Guillou's notes page 2, he wrote

$$ (X \vee Y)\wedge Z \cong (X \wedge Z) \vee (Y \wedge Z)$$

I could not find any reference for this fact. Bert Guillou works in the category of compactly generated weakly Hausdorff space (CGWH).

I also wonder what conditions are required for htis to hold in TOP.

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My thoughts: We have a homeomoprhism $$ (X \sqcup Y) \times Z \cong (X \times Z) \sqcup (Y \times Z)$$

We know the map $$ X \sqcup Y \times Z \rightarrow (X \vee Y) \wedge Z$$ is a quotient map. It then suffices to show that $$ (X \times Z) \sqcup (Y \times Z) \rightarrow (X \wedge Z) \sqcup (Y \wedge Z)$$ is a quotient map. But is this true?

EDIT: I think coproducts do respect quotients as a bifunctor. Specifically, we consider the UMP of $X_1 \sqcup X_2 \rightarrow Y_1 \sqcup Y_2$, wth a test space $Y_1 \sqcup Y_2 \rightarrow Z$.

Answer 1

I think that this works in general without any assumptions in Top.

Let $X$, $Y$ and $Z$ be pointed topological spaces, where we think of the wedge of two spaces $X \vee Y$ as a subspace of the product $X \times Y$.

There are natural maps in both directions. Define the map $e: X \wedge (Y \vee Z) \rightarrow (X \wedge Y) \vee (X \wedge Z)$ by: $\forall x \in X$,

\begin{cases}(x,(y,*)) \mapsto ((x,y),*) \; &\forall y \in Y \\ (x,(*,z)) \mapsto (*,(x,z)) \; &\forall z \in Z. \end{cases}

This map is well defined, and is continuous by the pasting lemma.

The map in the reverse direction does the obvious thing, and is similarly continuous

Answer 2

Update: Thanks to @Ken for pointing out my mistake in the comments. Here is a corrected answer.

If you want the distribution to be up to homeomorphism, then the smash product distributes over finite wedges for arbitrary based spaces. This doesn't work the same for infinite wedges in the regular topological category. However, in the general case, there is a canonical continuous bijection: $f:\bigvee_{j\in J}X\wedge Y_j\to X\wedge \bigvee_{j\in J}Y_j$. This bijection is a weak homotopy equivalence but may not always be a true homeomorphism. Algebraic topologists often deal with this difficulty by working internal to a convenient category such as the category of $k$-spaces.

Proof of distribution over finite wedges. Let $(X,x_0)$ and $(Y_j,y_j)$, $j\in J$ be based spaces where $J$ is finite. Let $w_0$ be the wedge point of $\bigvee_{j\in J}Y_j$. Consider the following commutative diagram of canonical quotient maps. $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccc} X\times \coprod_{j\in J}Y_j & \ra{\cong} & \coprod_{j\in J}X\times Y_j \\ \da{} && \da{} \\ X\times \bigvee_{j\in J}Y_j & \ra{p} & \bigvee_{j\in J}X\times Y_j\\ \da{} && \da{\vee_{j}q_j} \\ X\wedge \bigvee_{j\in J}Y_j & & \bigvee_{j\in J}X\wedge Y_j \end{array} $$

The top map is clearly a homeomorphism and we can identify these two spaces if we like. The middle horizontal map $p$ collapses $X\times \{w_0\}$ to the wedge point of $\bigvee_{j\in J}X\times Y_j$. If $q_j:X\times Y_j\to X\wedge Y_j$ is the canonical quotient, then the vertical map $\vee_{j}q_j:\bigvee_{j\in J}X\times Y_j\to \bigvee_{j\in J}X\wedge Y_j$ is quotient because wedges of based quoitient maps are quotient. Since all the maps are quotient maps, the universal property of quotient spaces tells you that have have the desired homeomorphism $X\wedge \bigvee_{j\in J}Y_j\cong \bigvee_{j\in J}X\wedge Y_j$ that makes the diagram commute if it holds that the two vertical compositions make the same identifications. If you check both of them, all fibers are points except that the fiber over each basepoint is $\left(X\times\coprod_{j\in J}\{y_j\} \right)\sqcup \left(\{x_0\}\times\coprod_{j\in J} Y_j\right)=\coprod_{j\in J}(X\times \{y_j\}\cup \{x_0\}\times Y_j)$. $\square$

Note: the intermediate quotient maps are not necessary to the proof but might help you keep track of things by breaking up the construction into steps.

In the general case (when $J$ is infinite), it's best to assume the spaces involved are Hausdorff. The right composition in the above diagram is quotient, but the left composition might not be quotient because the wedge $\bigvee_{j}Y_j$ will not always be locally compact Hausdorff. Hence, we get a continuous bijection $f:\bigvee_{j\in J}X\wedge Y_j\to X\wedge \bigvee_{j\in J}Y_j$ but this may not be a homeomorphism. However $f$ maps any compact subspace $K$ of $\bigvee_{j\in J}X\wedge Y_j$ homeomorphically onto $f(K)$ (we're using the closed mapping theorem here, which is why we want the spaces to be Hausdorff). This implies that $f$ is actually a bijective weak homotopy equivalence. In fancier categorical terms, it means that the two spaces have homeomorphic $k$-space coreflections, $k(\bigvee_{j\in J}X\wedge Y_j)\cong k(X\wedge \bigvee_{j\in J}Y_j)$.

Answer 3

I know this is an old question, but just want to point out that this is not special to the category of spaces, but to any context in which pushouts commutes with product (e.g. any category which admits an exponential object, such as the category of locally compact Hausdorff spaces, or any category in which pushouts are universal in the sense of this definition).

Given that, there is a canonical isomorphism $$A\times \left(B \lor C\right) := A \times \left(B \cup_* C\right) \simeq \left(A \times B \right)\cup_A\left(A\times C\right)$$ In addition there is always an obvious isomorphism $$A \lor B \lor C \simeq \left(A \lor B \right) \cup_A \left(A \lor C \right)$$ Now by definition $$A \land \left(B \lor C \right):= \operatorname{cof} \left(A \lor B \lor C \to A\times \left(B \lor C\right)\right)$$ Composing the last equation with the previous two isomorphisms, we get $$A \land \left(B \lor C \right) \simeq \operatorname{cof} \left( \left(A \lor B \right) \cup_A \left(A \lor C \right) \to \left(A \times B \right)\cup_A\left(A\times C\right)\right)$$ Finally, pushouts commute with pushouts, so the right term is canonically isomorphic to $$\left(A \land B\right) \lor \left(A \land C \right)$$