In algebraic topology wedge sum is defined for a family of indexed, say by a set $\mathcal{I}$, spaces $X_i$ with points $p_i \in X_i$ as: $$ \bigvee_{i \in \mathcal{I}} X_i = \frac{\bigsqcup_{i \in \mathcal{I}} X_i}{(i,p_i)\sim(j,p_j)}. $$

Exactly the same definition will work just for the pointed sets. But what happens then $\mathcal{I} = \emptyset$?

One resolution comes form the fact that wedge sum behaves as coproduct in the category of pointed topological spaces $\mathsf{TOP}^*$, or equivalently pointed sets $\mathsf{SET}^*$, for a reasonable (meaning $\mathcal{I} \neq \emptyset$) families of objects . Then in case of $\mathcal{I} = \emptyset$ the coproduct, if it exists, must be equal to the initial object of the corresponding category. So, in case with $\mathsf{SET}^*$, the zero object, so
$$ \bigvee_{i \in \emptyset} X_i = \Big(\{p\}, p \Big), $$ and the same resolution works for $\mathsf{TOP}^*$.

Is this a standard definition? Is there any reasons not to use it? Do algebraic topologists have any other conventions.

The only thing that worries me here is that together with the original elementary definition the resolution suggests a weird identity

$$ \frac{\emptyset}{\emptyset} = \bigvee_{i \in \emptyset} X_i = \{p\} $$

  • $\begingroup$ Do you mean the wedge-product $\vee$ or the smash-product $\wedge$? Note that the latex-command for wedge is not wedge but vee… $\endgroup$ Jul 4, 2021 at 17:48
  • $\begingroup$ Yeas I meant wedge-products. Sorry for the typo $\endgroup$
    – Nik Pronko
    Jul 4, 2021 at 17:49
  • $\begingroup$ Wedge sum, please :-) $\endgroup$ Jul 4, 2021 at 17:49
  • $\begingroup$ Right. Shame on me :) $\endgroup$ Jul 4, 2021 at 18:01

2 Answers 2


You are right, the empty wedge sum is the initial object of pointed spaces. The formula you suggested for the wedge sum is correct for non-empty index sets*. The formula which always works is $$\bigvee_{i \in I} (X_i,p_i) = \left(\coprod_{i \in I} X_i \sqcup\{p\} ~/~ (p_i \sim p),p\right).$$ There is a more conceptual explanation for this: Consider the forgetful functor $ \mathbf{Top}_* \to \mathbf{Top}$. It is monadic, the monad sends a space $ X$ to $X \sqcup \{\star\}$, the unit is the evident map $X \hookrightarrow X \sqcup \{\star\}$, and the multiplication is the evident map $X \sqcup \{\star\} \sqcup \{\star\} \to X \sqcup \{\star\}$. There is a general procedure how to produce colimits in $ \mathrm{Alg}(T)$ for a "well-behaved" monad $T$ (for example, it should preserve reflexive coequalizers, which is the case here) on a cocomplete category $\mathcal{C}$: Given a diagram of $T$-algebras, take the colimit of the underlying objects in $\mathcal{C}$, then apply $T$ to get the free $T$-algebra, and finally quotient out the relations which make sure that the inclusions become $T$-algebra maps. Notice that this is exactly what happened in the formula above: To get the coproduct of pointed spaces $(X_i,p_i)$, we first form the coproduct $\coprod_{i \in I} X_i$ of the underlying spaces, adjoin freely a base point $p$ to get the free pointed space on it, and finally identify each $p_i$ with $p$ in order to make sure that the inclusions from $X_i$ become pointed maps.

*It is also correct for the empty set with the correct definition of $X/A$ as the pushout of $\{\star\} \leftarrow A \hookrightarrow X$ as observed by PrudiiArca.


In any category the coproduct over an empty indexing set is the initial object (assuming existence of things), so you are right that in $\mathsf{Top}_\ast$ we have $$\bigvee \limits_{i\in\emptyset} (X_i, x_i) = (\{*\},*)$$

Your unintuitive identity doesn’t make sense to me, since $\emptyset$ is uninhabited / empty, hence cannot be regarded as a pointed topological space. But anyway you are right in the assumption that $$\emptyset/\emptyset=*$$ is a common convention, usually seen in the context of comparing reduced- and unreduced (co)homology theories…

  • $\begingroup$ What is your (general) definition of $X/A$ for a subspace $A$ of $X$? $\endgroup$ Jul 4, 2021 at 18:02
  • 2
    $\begingroup$ I would have said the coequalizer of the inclusion of $A$ with a constant map on any point of $A$, but I must admit I didn’t really think about it until now… I think if I use the pushout of the inclusion and the terminal map the definition is more general and specializes to $\emptyset/\emptyset = *$… Cool! $\endgroup$ Jul 4, 2021 at 19:00
  • $\begingroup$ That's cool. You may add it to mathoverflow.net/questions/45951 $\endgroup$ Jul 4, 2021 at 20:06
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    $\begingroup$ More generally, $X/\emptyset = X \sqcup \{\star\}$. $\endgroup$ Jul 4, 2021 at 20:11
  • $\begingroup$ Which is precisely the convention used in homology theory. This is nice, thanks! $\endgroup$ Jul 4, 2021 at 20:18

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