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While discussing categories without coproducts, we stumbled with the category $\mathbf{Met}$ that takes metric spaces as its objects and short maps as its morphisms. It is claimed that $\mathbf{Met}$ does not have coproducts. I'm puzzled since the disjoint union of metric spaces is also a metric space so that would make it a candidate for the coproduct in the category but I'm unable to see why it fails to be one. Any ideas?

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    $\begingroup$ In the disjoint union, what is the distance $d(x_1,x_2)$ for $x_1\in X_1$ and $x_2 \in X_2$? 't Seems nontrivial to me. $\endgroup$ – Lord_Farin May 9 '16 at 18:05
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An easy argument proceeds by contradiction: Suppose that there were coproducts, and consider $X+Y$ for nonempty $X,Y$. Now consider the metric spaces $2_N =\{-N,N\} \subseteq \Bbb R$ together with the (obviously short) maps $$\begin{align}+&:X \to 2_N, x \mapsto N\\-&:Y\to2_N,y\mapsto -N\end{align}$$

Then $d(+(x),-(y)) = 2N$ for arbitrary $x,y,N$; it is easy to see that this contradicts the simultaneous shortness of all induced maps $X+Y\to 2_N$ (finitely many go fine, but the distance between $x \in X$ and $y \in Y$ in $X+Y$ needs to be infinite to make them all short).

Therefore, the coproduct $X+Y$ cannot exist. The fundamental reason behind this is the inability to make the distance between $X$ and $Y$ large enough.

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  • $\begingroup$ If I got it right, then the key is to realize that the distance between points $x \in X$ and $y \in Y$ in $X+Y$, say $d(x,y)$, is finite so we can find some $N \in \mathbb{R}$ such that $N > d(x,y)$. With such an $N$, for the metric space $2_N$, it would be impossible to find a short map having the universal property for coproducts. Really compelling argument, thanks! $\endgroup$ – egomezcana May 9 '16 at 22:23
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    $\begingroup$ @egomezcana That is exactly right, and you're welcome :). $\endgroup$ – Lord_Farin May 9 '16 at 22:49
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EDIT: I wrote this answer a few seconds before Lord_Farin's answer appeared, so in the first sentence below I was referring to his comment on the question, not to his answer. I guess I'll keep this here for the comment about allowing $\infty$ as a distance, but regard this more as a comment now that he's posted a full proof.


@Lord_Farin hit on the difficulty. The easiest way to see what goes wrong is to try to make the coproduct of a point with itself: you might think it's two points, but at what distance? If the distance is $d$, the universal property fails for maps from this coproduct to any space with some pair of points at a distance greater than $d$.

If you change the definition of metric slightly to allow $d(x_1,x_2)=\infty$, then the resulting category of slightly generalized metrics spaces and short maps does have coproducts: in the coproduct of $X_1$ and $X_2$ you make the distance from points of $X_1$ to $X_2$ infinite.

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  • $\begingroup$ Thanks for the insight! It was very useful to understand the one provided by Lord_Farin, I just took $X=Y=\{\ast\}$ and everything ran smoothly. Also, I find very interesting that if you allow infinite distances, the category has coproducts. $\endgroup$ – egomezcana May 9 '16 at 22:35

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