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A. Helemskii in the book "Lectures on functional analysis" write (in my horrible translation):

The category of Hausdorff topological spaces (morphisms are continuous maps) contain the full subcategory of metrizable topological spaces. Note: metrizable, not metric; the category of metric spaces (morphisms are continuous maps) is not a subcategory of the category of topological spaces (such as the category of linear spaces is not a subcategory of the category of sets).

Why the category of metric spaces $\neq$ the category of metrizable topological spaces?

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    $\begingroup$ I believe it's because the category of metric spaces normally has a very restricted class of morphism (much more restrictive than continuous). I could be wrong, but I think it is normally restricted to the 'non-expansive maps'. Those $f\colon X\rightarrow Y$ such that $d_X(x,x')\geq d_Y(f(x),f(x'))$. $\endgroup$
    – Dan Rust
    Aug 6, 2013 at 17:22
  • $\begingroup$ @DanielRust: I've edit my question. Define morphisms in metric spaces as continuous maps. $\endgroup$ Aug 6, 2013 at 17:25
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    $\begingroup$ Note that there is a functor from Met (with continuous maps) to MTop (metrisable space) which only remembers the topology of a metric space, but in general there is no canonical inverse functor from MTop to Met because there may be more than one metric that can be put on a metrisable space. $\endgroup$
    – Dan Rust
    Aug 6, 2013 at 17:31
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    $\begingroup$ It seems I understand. The topology of a metrizable space encodes a class of equivalent metrics ('equivalent' mean generate the same topology). But does Met and MTop are equivalent categories? $\endgroup$ Aug 6, 2013 at 17:37
  • $\begingroup$ You are right in any case that the two categories are not EQUAL, but they may be equivalent. $\endgroup$ Aug 6, 2013 at 18:32

1 Answer 1

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A metric space is a pair $(X,d)$ consisting of a set $X$ and a metric $d$. In contrast, a metrizable space is topological space $(X,\tau)$ such that there is a metric $d$ which induces the topology $\tau$. This metric $d$ may not be unique.

There is a faithful functor $U:\mathbf{Met}\to\mathbf{MTop}$ which send $(X,d)$ to $(X,\tau)$ and a map $f$ to itself. Depending on what morphisms $\mathbf{Met}$ has, this may or may not be a full functor. If $U$ is not a full functor, then $\mathbf{Met}$ cannot be equivalent to $\mathbf{MTop}$.
On the other hand, if you take all the continuous maps as the morphisms in $\mathbf{Met}$, then $U$ is fully faithful. Since for each metrizable space $(X,\tau)$ there is a metric space $(X,d)$ with $U(X,d)=(X,\tau)$, $U$ is also surjective, which implies that it is part of an adjoint equivalence $(F,U;\text{Id}_{(X,\tau)},\varepsilon)$. As we see, the unit $\eta$ is the identity on the space $(X,\tau)$. For such adjunctions, the left adjoint $F$ is called a left-adjoint-right-inverse (The "-right-inverse" comes from the fact that if $\eta$ is the identity, then $UF$ is the identity functor, so $F$ is a right inverse to $U$.)

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