0
$\begingroup$

I started with Rotman Algebraic Topology, that is Exercises 0.11 from the zeroth chapter, which introduces basic category theory.

here is the definition for subcategory in Rotman:

Rotmans definition of subcategory

I find his definition of subcategory a little bit confusing, so i will use the definition from wikipedia. (https://en.wikipedia.org/wiki/Subcategory) to come up with a strategy to show that $Top$ is a subcategory of $Top^2$

Show that one may regard Top as a subcategory of $Top^2$ if one identifies a space X with the pair $(X,\varnothing)$.

possible Strategy?

1) Show for every X in obj(Top), the identity morphism $1_X \in Hom_{Top}$

2) for every morphism $f : X \to X \in Hom_{Top}$, both the source X and the target X are in $obj(Top)$

3) for every pair of morphisms f and g in $Hom_{Top}$ the composite $f \circ g$ is in $Hom_{Top}$ whenever it is defined.

i am bit lost how to show that $(X, \varnothing)$ is really a subcategory of $Hom_{Top^2}$. How to start?

  • 2
    Any object $X$ in $Top$ can be identified with $(X,\emptyset)$ and any morphism $X\to Y$ can be identified with a morphism $(X,\emptyset)\to (Y,\emptyset)$ in $Top^2$. I guess that's all (assuming that $Top^2$ is the category of topological pairs) – 2017-02-18
  • 0
    What is $\mathrm{Top}^2$? – 2017-02-18
  • 0
    @coconut obj $Top^2$ are the ordered pairs (X,A), where A is a subspace of X. so when i set $(X,\varnothing)$ this should work, but i dont see it. – 2017-02-18
  • 0
    @PeterFranek i think i got it :) Really Just Assume the ordered pair (X, \varnothing) is an object in Top^2, same for the morphisms. then i can show that Top forms the subcategory with the composition, same morphism, etc. – 2017-02-18

0 Answers 0