Top.yaml

id: Top
name: category of topological spaces
notation: $\Top$
objects: topological spaces
morphisms: continuous functions
description: This is the most basic category of geometric objects.
nlab_link: https://ncatlab.org/nlab/show/Top
dual_category: Top_op

tags:
  - topology

related_categories:
  - Haus
  - Met_c
  - Top*

satisfied_properties:
  - property: locally small
    proof: There is a forgetful functor $\Top \to \Set$ and $\Set$ is locally small.

  - property: complete
    proof: Take the limit of the underlying sets and endow it with the coarsest topology making all projections continuous.

  - property: cocomplete
    proof: Take the colimit of the underlying sets and endow it with the finest topology making all inclusions continuous.
    check_redundancy: false

  - property: well-powered
    proof: This is clear from the classification of monomorphisms as injective continuous maps.

  - property: well-copowered
    proof: This is clear from the classification of epimorphisms as surjective continuous maps.

  - property: semi-strongly connected
    proof: Every non-empty space is weakly terminal (by using constant maps).

  - property: generator
    proof: The one-point space is a generator since it represents the forgetful functor $\Top \to \Set$.

  - property: cogenerator
    proof: It is easily checked that the indiscrete two-point space is a cogenerator.

  - property: infinitary extensive
    proof: '[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow $Y_i$ with the subspace topology. If $f$ is continuous, each $Y_i = f^{-1}(X_i)$ is open in $Y$, so that $Y = \coprod_i Y_i$ holds as topological spaces, and each $f_i$ is continuous.'

  - property: regular subobject classifier
    proof: The indiscrete two-point space $\{0,1\}$ is a regular subobject classifier since continuous maps $X \to \{0,1\}$ correspond to subsets of $X$.

  - property: coregular
    proof: The category has all limits and colimits, and the regular monomorphisms are the subspace inclusions. Thus, it suffices to prove that subspace inclusions are stable under pushouts. For a proof see e.g. Lemma 3.6 at the <a href="https://ncatlab.org/nlab/show/subspace+topology#pushout" target="_blank">nLab</a>.

  - property: filtered-colimit-stable monomorphisms
    proof: This follows from Lemma 2 <a href="/content/subcategories">here</a> applied to the forgetful functor to $\Set$.

unsatisfied_properties:
  - property: skeletal
    proof: This is trivial.

  - property: balanced
    proof: If $X$ is a set, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X$ has at most one element.

  - property: cartesian filtered colimits
    proof: 'The functor $\IQ \times - : \Top \to \Top$ does not preserve sequential colimits, see <a href="https://math.stackexchange.com/questions/1255678" target="_blank">MSE/1255678</a>.'

  - property: regular
    proof: See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>.

  - property: accessible
    proof: In fact, it does not have any small colimit-dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>. For a related result, see <a href="https://mathoverflow.net/questions/288648" target="_blank">MO/288648</a>.

  - property: coaccessible
    proof: 'Assume $\Top$ is coaccessible. Let $p : S \to I$ be the identity map from the Sierpinski space to the two-element indiscrete space. Then, a topological space is discrete if and only if it is projective to the morphism $p$. This implies that the full subcategory spanned by all discrete spaces, which is equivalent to $\Set$, is coaccessible by Prop. 4.7 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. However, since <a href="/category/Set">$\Set$</a> is not coaccessible, this is a contradiction.'

  - property: co-Malcev
    proof: 'See <a href="https://mathoverflow.net/questions/509548" target="_blank">MO/509548</a>. We can also phrase the proof as follows: Consider the forgetful functor $U : \Top \to \Set$ and the relation $R \subseteq U^2$ defined by $R(X) \coloneqq \{(x,y) \in U(X)^2 : x \in \overline{\{y\}} \}$. Both are representable: $U$ by the singleton and $R$ by the Sierpinski space. It is clear that $R$ is reflexive, but not symmetric.'

  - property: cofiltered-limit-stable epimorphisms
    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Top$ which equips a set with the indiscrete topology.

  - property: effective cocongruences
    proof: 'Consider the indiscrete topological space $I$ on two points. This represents the functor which takes a topological space $X$ to the pairs of indistinguishable points of $X$. Therefore, we get a cocongruence $1 \rightrightarrows I$, where the maps are the two possible functions. However, this cannot be effective: if we have $h : Z\to 1$ which equalizes the two maps, then $Z$ must be empty. But that means the cokernel pair of $h$ is the discrete space on two points.'

special_objects:
  initial object:
    description: empty space
  terminal object:
    description: singleton space
  coproducts:
    description: disjoint union with the disjoint union topology
  products:
    description: direct product with the <a href="https://en.wikipedia.org/wiki/Product_topology" target="_blank">product topology</a>

special_morphisms:
  isomorphisms:
    description: homeomorphisms
    proof: This is easy.
  monomorphisms:
    description: injective continuous maps
    proof: For the non-trivial direction, the forgetful functor to $\Set$ is representable (by the terminal object), hence preserves monomorphisms.
  epimorphisms:
    description: surjective continuous maps
    proof: The proof works exactly as for the category of sets, where we endow $2$ with the indiscrete topology.
  regular monomorphisms:
    description: embeddings
    proof: 'Equalizers are embeddings by their construction. Conversely, if $f : X \to Y$ is an embedding, then $f$ is the equalizer of the two characteristic maps $\chi_Y, \chi_{f(X)} : Y \to \{0,1\}$, where $\{0,1\}$ carries the indiscrete topology.'
  regular epimorphisms:
    description: surjective quotient maps
    proof: 'Regular epimorphisms are surjective quotient maps by the explicit construction of coequalizers. Conversely, if $q : X \to Y$ is a surjective quotient map, then one checks that $q$ is the coequalizer of its kernel pair $X \times_Y X \rightrightarrows X$: This is true for the underlying sets, and continuity of the induced morphism follows since $q$ is a quotient map.'