Well, first of all, does the coproduct exist in the category of path-connected spaces, and if not how would you prove it? If it does exist what is it and how do you find it? The usual coproduct of topological spaces is disjoint union, but it is not path-connected.

As a follow up question, given a full subcategory of a category in which products/coproducts do exist, what is the strategy for finding out if that category also has products/coproducts?


They don't exist. For instance, suppose there existed a coproduct $Y=X\coprod X$, where $X$ is a point. Then there would be a unique map $f:Y\to [0,1]$ sending the first copy of $X$ to $0$ and the second copy of $X$ to $1$. Since $Y$ is path-connected, $f$ must be surjective. But now let $g:[0,1]\to[0,1]$ be any non-identity map that sends $0$ to $0$ and $1$ to $1$. Then $gf$ is another map $Y\to[0,1]$ sending the first copy of $X$ to $0$ and the second copy of $X$ to $1$, and $gf\neq f$ since $f$ is surjective. This is a contradiction.

In general, when trying to find limits/colimits in a full subcategory, you are trying to find the "canonical approximation" to the (co)limit in the ambient category which lies in your subcategory. So if the (co)limit in the ambient category lies in your subcategory, it is the (co)limit. If it doesn't, you try to modify it to lie in your subcategory in a "universal" way (more precisely, such that maps between the (co)limit and objects of your subcategory are in bijection with maps between the modification and objects in your subcategory). If there doesn't seem to be any sort of universal way to modify it, that's a good sign that the (co)limit doesn't exist in your subcategory. For instance, in this case, there doesn't seem to be any particularly "universal" way to take a disjoint union of two path-connected spaces and make it path-connected. You can then try to make some concrete argument that it doesn't exist like the one above; the details of how you do this will depend a lot on what your category is.

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  • $\begingroup$ Would it work in the category of connected spaces? $\endgroup$ – AIM_BLB Dec 19 '18 at 21:05
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    $\begingroup$ The exact same argument works for connected spaces (you can still conclude that $f$ is surjective from $Y$ being connected). $\endgroup$ – Eric Wofsey Dec 19 '18 at 21:07
  • $\begingroup$ Right! Makes sense. Thanks :) $\endgroup$ – AIM_BLB Dec 19 '18 at 21:29

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