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Let $X$ and $Y$ be two topological spaces and let $A \subseteq X$ and $B \subseteq Y.$ Then $$X/A \wedge Y/B \cong (X \times Y)/ ((X \times B) \cup (A \times Y)).$$

First we need to get a continuous surjective map $f : X \times Y \longrightarrow X/A \coprod Y/B$ and if we compose it with the quotient map $p : X/A \coprod Y/B \longrightarrow X/A \wedge Y/B$ then we get a continuous surjective map $g : X \times Y \longrightarrow X / A \wedge Y / B.$ Now define an equivalence relation on $X \times Y$ as follows $:$ $$(x,y) \sim (x',y') \iff g(x,y) = g(x',y').$$ Then by universal property of quotient topology there exists a continuous surjective map $\overline g : (X \times Y)/ \sim \longrightarrow X/A \wedge Y/B.$ Furthermore, $\overline g$ is a homeomorphism iff $g$ is a quotient map. This is the so called theory.

Now the questions are $:$

$(1)$ How to define $f$ so that the set of fibres of $g : = (p \circ f)$ is $(A \times Y) \cup (X \times B)\ $?

$(2)$ For such an $f$ how to show that $g$ is a quotient map?

Would anybody please help me in this regard? Thanks in advance.

ACB
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  • What is the origin of your question? A textbook? In general it is not true, you need additional conditions on the pairs $(X,A), (Y,B)$. – Paul Frost Apr 18 '21 at 07:39
  • It was given by our instructor @Paul Frost. BTW do you know any video lectures where things like that are being covered? The way our instructor introduced such concepts is very abstract. But algebraic topology is useless to study without geometric intuition. – ACB Apr 18 '21 at 07:45
  • No, I do not know such video lectures. – Paul Frost Apr 18 '21 at 08:32

1 Answers1

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There is a continuous bijection $$\phi : (X \times Y)/ ((X \times B) \cup (A \times Y)) \to X/A \wedge Y/B .$$ To see this, consider the the map $$\psi : X \times Y \stackrel{p_X \times p_Y}{\longrightarrow} X/A \times Y/B \stackrel{q}{\longrightarrow} X/A \wedge Y/B $$ where $p_X, p_Y, q$ are the obvious quotient maps. It is easy to verify that the fibers $\psi^{-1}(\xi)$ are singletons if $\xi \ne *$ (where $*$ is the point obtained by collapsing $X/A \times \{*\} \cup \{*\} \times Y/B$) and $\psi^{-1}(*) = X \times B \cup A \times Y$. Hence we get $\phi$.

Now $\phi$ is a homeomorphism if and only if $\psi$ is a quotient map. The problem is that the product of quotients maps $p_X \times p_Y : X \times Y \to X/A \times Y/B $ is not always a quotient map. Make a search in this forum for "product quotient maps". There are many hits, for example When is the product of two quotient maps a quotient map?

Thus you will need additional assumptions on the pairs $(X,A), (Y,B)$.

Update:

The fibers of $q : X/A \times Y/B \to X/A \wedge Y/B = (X/A \times Y/B)/(X/A \times \{*\} \cup \{*\} \times Y/B)$ are singletons for $\xi \ne *$ and $q^{-1}(*) = X/A \times \{*\} \cup \{*\} \times Y/B$.

Clearly $$(p_X \times p_Y)(X \times B \cup A \times Y) = (p_X \times p_Y)(X \times B) \cup (p_X \times p_Y)(A \times Y) \\= p_X(X) \times p_Y(B) \cup p_X(A) \times p_Y(Y) = X/A \times \{*\} \cup \{*\} \times Y/B .$$ Moreover $$(X \times Y) \setminus (X \times B \cup A \times Y) = ((X \times Y) \setminus (X \times B)) \cap (X \times Y) \setminus (A \times Y)) \\ = X \times (Y \setminus B) \cap (X \setminus A) \times Y = (X \setminus A) \times (Y \setminus B) .$$

But $p_X \times p_Y$ is injective on $(X \setminus A) \times (Y \setminus B)$ and for $x \in X \setminus A$ resp. $y \in Y \setminus B$ we have $p_X(x) \ne *$ resp. $p_Y(y) \ne *$. Thus $$(p_X \times p_Y)((X \times Y) \setminus (X \times B \cup A \times Y)) \subset (X/A \times Y/B) \setminus (X \times * \cup * \times Y/B).$$ We conclude that $(p_X \times p_Y)^{-1}(X/A \times \{*\} \cup \{*\} \times Y/B) = X \times B \cup A \times Y$ and that for $\eta \notin X/A \times \{*\} \cup \{*\} \times Y/B$ the fiber $(p_X \times p_Y)^{-1}(\eta)$ is a singleton.

This shows $\psi^{-1}(*) = X \times B \cup A \times Y$. For $\xi \ne *$ the fiber $q^{-1}(\xi)$ a singleton outside of $X/A \times \{*\} \cup \{*\} \times Y/B$, thus $\psi^{-1}(\xi)$ is a singleton.

Paul Frost
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  • Can you show the computation as to how to compute pre-images of $\xi \neq \star$ and $\star$ under the mapping $\psi\ $? I find it difficult to compute it. Thanks. – ACB Apr 18 '21 at 08:48
  • I have computed it on my own and found that the inverse image of $\star$ under $\psi$ is depending upon whether or not the base points $\star$ belong to the spaces $A$ and $B$ which the spaces are quotienting with. There are four options for the inverse image of $\star$ under the map $\psi$ (depending on whether or not the base points $\star$ belongs to $A$ or $B$) which are $:$ $$\begin{align}(1)\ (X \times B) \cup (A \times Y) \ (2)\ (X \times {\star}) \cup (A \times Y) \ (3)\ (X \times A) \cup (Y \times {\star}) \ (4)\ (X \times {\star}) \cup (Y \times {\star}) \end{align}$$ – ACB Apr 18 '21 at 16:34
  • @Phibetakappa $X,Y$ do not have basepoints. Basepoints (denoted as $*$) are generated by forming the quotients $X/A, Y/B$. This is why we can form the smash product $X/A \wedge Y/B$. – Paul Frost Apr 18 '21 at 17:21
  • So the basepoints of $X/A$ and $Y/B$ are precisely the images of $A$ and $B$ under the respective quotient maps from $X$ and $Y$ onto $X/A$ and $Y/B.$ Is that what you mean? – ACB Apr 18 '21 at 17:33
  • Yes, that is correct. – Paul Frost Apr 18 '21 at 21:32