Consider the Category $C$ of topological spaces with Homotopy classes of continous maps as Morphisms.

An Object $T$ is terminal if for every object $X\in C$ there exist a single morphism $X\rightarrow T$.

I believe that there are no such objects because there is no homotopy (in general) from an arbitrary topological space to, for example, a nullhomotopic space. I amconfused and would appreciate some insights.

  • 2
    $\begingroup$ homotopy classes of maps as morphisms? $\endgroup$ Sep 19, 2018 at 0:27
  • $\begingroup$ Yes. I will add that to the question, sorry. $\endgroup$ Sep 19, 2018 at 1:13

2 Answers 2


I think you're making the mistake of thinking the maps are homotopy equivalences.

If $T$ is contractible, then for most spaces $X$, there indeed does not exist a homotopy equivalence $X \to T$.

However, given any two nonempty spaces $X$ and $Y$, there always exists a continuous map $f : X \to Y$ (e.g. a constant function). The homotopy class of $f$ is therefore an element of $\hom_{\mathbf{hTop}}(X, Y)$.

In particular, if $Y$ is a one-point space, then there is a unique continuous map $X \to Y$ (even if $X$ is empty). Thus there is a unique homotopy class of maps $X \to Y$, and so $Y$ is terminal in hTop.

If $T$ is homotopy equivalent to $Y$, then it is isomorphic to $Y$ in hTop, so $T$ is terminal as well.

This argument can be generalized:

Theorem: If $F : \mathcal{C} \to \mathcal{D}$ is a full and essentially surjective functor and $T$ is a terminal object of $\mathcal{C}$, then $F(T)$ is a terminal object of $\mathcal{D}$.

  • $\begingroup$ thank you for the generalization! $\endgroup$ Sep 19, 2018 at 2:52

Every map $f:X \to \{pt\}$ is continuous, and homotopic to a constant map. It follows that the object $\{pt\}$ is terminal.

In particular, there is always a constant map $X \to \{pt\}$ and it is clearly unique.


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