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Suppose the object of the category are metric spaces and for $\left(A,d_A\right)$ and $\left(B,d_B\right)$ metric spaces over sets A and B, a morphisms of two metric space is given by a function between the underlying sets, such that $f$ presere the metric structure: $\forall x,y,z \in A$ we have:

  • $$ d_B\left(f\left(x\right),f\left(y\right)\right)= 0 \Leftrightarrow f\left(x\right)=f\left(y \right)$$
  • $$d_B\left(f\left(x\right),f\left(y\right)\right)=d_y\left(f\left(y\right),f\left(x\right)\right)$$
  • $$d_B\left(f\left(x\right),f\left(y\right)\right) \le d_B\left(f\left(x\right),f\left(z\right)\right) + d_B\left(f\left(z\right),f\left(y\right)\right) $$ and furthermore : $\forall \epsilon > 0$, $\exists \delta >0 $ which satisfy:
  • $$d_A\left(x,y\right)<\delta \Rightarrow d_B \left(f\left(x\right),f\left(y\right)\right)< \epsilon$$

Is this the category of metric spaces and continues functions? What if we drop the last requirement?

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The first three axioms are redundant; they're already implied by the fact that $(B, d_B)$ was assumed to be a metric space. The fourth axiom is continuity. –  Qiaochu Yuan Nov 3 '12 at 17:52
    
@ Qiaochu Yuan: Is that continuity or uniform continuity? –  Hooman Nov 3 '12 at 18:47
    
My apologies. Yes, it's uniform continuity. –  Qiaochu Yuan Nov 3 '12 at 20:39

1 Answer 1

I don't think there's really one the category of metric spaces. The fourth axiom here gives you a category of metric spaces and (uniformly) continuous functions. The other axioms are implied by the assumptions. Allowing $\delta$ to depend on $x$ gives you the category of metric spaces and (all) continuous functions.

One way to preserve metric structure would be to demand that $d_B(f(x),f(y))=d_A(x,y)$. This would restrict the functions to isometric ones, which are all homeomorphisms, so you could relax the restriction to $d_B(f(x),f(y))\le d_A(x,y)$. That way you get the category of metric spaces and contractions.

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