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Let $(X,x_0)$ be a pointed space with base point $(x_0)$. Let $\bigvee_{j \in J} X_j$ be the quotient of $\coprod_{j \in J} X_j$ where all bases are identified to a new base point.

This is the definition provided in the book. The problem is as follows:

Let $(X_j,x_j \mid j \in J)$ be a family of pointed spaces. Let $\bigvee_j^{'} X_j$ be the subset of those points $(a_j)$ $\in \prod_j X_j$ where all but one $a_j$ are equal to the base point. There is a canonical bijective continuous map $\bigvee_j X_j$ $\rightarrow$ $\bigvee_j^{'} X_j$. If $J$ is finite then this map is a homeomorphism. If $J$ is infinite and $(X_j,x_j) = (I,0)$, then it is not a homeomorphism

Unfortunately I have no idea where to begin and also I have a question regarding the text in bold, what is the one $(a_j)$ which is not the base point, are the base points already identified here? Regards

Paul Frost
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Sam Hoffman
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    There is no identification going on, $\prod_jX_j$ is a product and its elements are tuples $(a_j){j\in J}$. The subspace $\bigvee{j\in J}^{\prime}X$ is the one consisting of the tuples $(a_j)_{j\in J}$ for which there exists an index $j_0\in J$ such that $a_j=x_j$ for all $j\neq j_0$. That's what "all but one being equal to the basepoint" means (the "one" here is $j_0$). – Thorgott Mar 12 '22 at 12:42
  • In your first definition you also need a family of pointed spaces as in the second definition. – Paul Frost Mar 14 '22 at 17:14

2 Answers2

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For both $\bigvee$ and $\bigvee'$ you need the same family of pointed spaces $(X_j,x_j)$. In your first definition it does not make sense to consider a single pointed $(X,x_0)$.

  1. Define $f_k : X_k \to \bigvee' X_j, f_k(a_k) =$ point with coordinate $a_k$ for $j = k$ and coordinate $x_j$ else. Since $\bigvee' X_j$ is a subspace of $\prod X_j$, all $f_k$ are continuous (because all coordinate functions are continuous). All $x_k$ are mapped by the $f_k$ to the basepoint $(x_j) \in \bigvee' X_j$, thus the $f_k$ induce a continuous $F : \bigvee X_j = \left(\coprod X_j\right)/(x_j \sim x_k) \to \bigvee' X_j$. This map is clearly a bijection.

  2. For finite $J$ the map $F$ is a homeomorphism. This is proved for two factors in $X\vee Y\cong (X\times\{y_0\})\cup (\{x_0\}\times Y)$. You can easily generalize this to finitely many factors.

  3. If $J$ is infinite and $(X_j,x_j) = (I,0)$, then $F$ is not a homeomorphism: Let $U_j = [0,1/2)$ which is open in $I$. Then $\bigvee U_j$ is open in $\bigvee X_j$ because its preimage in $\coprod X_j$ is open. But $F(\bigvee U_j)$ is not open in $\bigvee' X_j$. If it were open, then it would have the form $F(\bigvee U_j) = V \cap \bigvee' X_j$ with an open $V \subset \prod X_j$ which contains $(x_j)$. Any such open $V$ must contain a set of the form $\prod V_j$ where $x_j \in V_j$ and all but finitely many $V_j = X_j = I$. This is impossible since $F(\bigvee U_j)$ does not contain any point with a coordinate $=1$.

Paul Frost
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That identification is quite handy when defining the smash products of spaces, as $X\wedge Y = X\times Y/X\vee Y$ can be thought of an actual quotient space by identifying $X\vee Y$ with $\{ (x, y_0)\in X\times Y\}\cup\{(x_0, y)\in X\times Y\}$, where $x_0\in X, y_0\in Y$ are the basepoints. Now, the idea here is similar, you take the sets $X'_j=\{ (x_j)_{j\in\mathscr{J}}\in \prod_{j\in\mathscr{J}}X_j\ |\ x_j\in X_j, x_i=*_i\text{ for }i\neq j\}$ with $*_i\in X_i$ being the basepoint. Then, for any $j\in\mathscr{J}$, $X'_j\approx X_j$, and the intersection is easy to find: $\bigcap_{j\in\mathscr{J}}X_j'=\{ (*_j)_{j\in\mathscr{J}}\}$. You can verify quickly that this general thing coincides with what I wrote first if $\mathscr{J}$ consists of two elements. The space $\bigvee'_{j\in\mathscr{J}}X_j$ is now the union of the spaces $X_j'$.

To construct a homeomorphism, take any $X_j$ and send every point $x_j$ to the point in $X_j'$ with the $j$th coordinate $x_j$ (and others $x_i = *_i$). This is one direction. The other direction takes a bit of checking, but you can do it using the first paragraph I wrote. Good luck!