In the image below, I am unable to understand the following points in Lemma 1.5.1:

1. Why do we need $C \times 2$ as the domain of functor H?
2. What does "$H$ restricts along $i_0$ and $i_1$ to the functors $F$ and $G$" mean?
3. How do we get a bijection between a natural transformation and functor $H$?

Equivalence of Categories

  • $\begingroup$ Can you provide some background knowledge? Where did you get the lemma? What is your background knowledge? Without this information it could be difficult providing a satisfying answer. $\endgroup$ Commented Apr 8, 2018 at 14:49
  • $\begingroup$ The lemma is from Section 1.5 "Equivalence of Categories" from "Category Theory in Context" by Emily Riehl. $\endgroup$
    – cbro
    Commented Apr 9, 2018 at 4:06

2 Answers 2


I will assume you are familiar with homotopies between continuous maps. If not, I suggest you try to understand that concept first and then try to understand this analogy.

To talk about the analogy let $X,Y$ be topological spaces and $f,g\colon X\to Y$ continuous maps. Let $F\colon X\times I\to Y$ be an homotopy between them, that is a continuous map such that $F|_{X\times \{ 0\}}=f$ and $F|_{X\times \{ 1\}}=g$.

  1. In the analogy, just as before we had $X\times I$ as the domain of our homotopy $F$, now we will have $C\times \mathbb{2}$ as the domain of our "homotopy-functor" $H$.

  2. This means precisely what is written in the diagram below, and the word restricts is justified precisely by this analogy, in which $F$ does really restrict to $f$ on $X\times \{ 0\}$ and to $g$ on $X\times \{ 1\}$. Note that in the topological case you can also express this with a very similar diagram, namely using bottom and top inclusions $i_{0},i_{1}\colon X \to X\times I$.

  3. The bijection is a nice and easy exercise that I don't want to spoil. Just write down what everything is and you will see how to assign to each natural transformation $\alpha$ such a functor $H$ and vice versa. Hints: objects are easy. For morphisms, note that there are 3 cases: $f\times \rightarrow $, $f\times id_{0}$ or $f\times id_{1}$ for some morphism $f$ in $C$. The last two cases are easy because you know that they must be the corresponding functor $F$ or $G$. For the first case use (resp. define) the natural transformation through its component at the domain of $f$ and then compose with the morphism corresponding to the naturality condition. To make it more intuitive, think of the map $f\times \rightarrow \colon (c_{1},0) \to (c_{2},1)$ as the composition $$(c_{1},0)\to (c_{1},1) \to (c_{2},1)$$

  • $\begingroup$ Thanks for the Homotopy analogy! I didn't think about it that way. But now it makes sense. $\endgroup$
    – cbro
    Commented Apr 9, 2018 at 4:04

I see that this question has an already accepted answer, nevertheless allow me to add something here.

  1. What does "$H$ restricts along $i_0$ and $i_1$ to the functors $F$ and $G$" mean?

It means that the composites $H \circ i_0$ and $H \circ i_1$ are equal to $F$ and $G$ respectively (that is the two triangles in the diagram commutes).

  1. Why do we need $C \times 2$ as the domain of functor $H$?

That is probably the hardest point to answer. Observe that if the $H$ restricts to $F$ and $G$ through the $i_j$'s then for each object $c \in C$ you have the morphism $$(c,0) \longrightarrow (c,1)$$ whose image through $H$ provide you with a morphism $$F(c)=H(c,0) \longrightarrow H(c,1)=G(c)$$ whose source and target coincie with those of the $c$-component of a natural transformation.

On the other hand for every $f \in C[c,c']$ you have a commutative square $$\require{AMScd}\begin{CD} (c,0) @>>> (c,1) \\ @V(f,\text{id}_0)VV @VV(f,\text{id}_1)V \\ (c',0) @>>> (c',1) \end{CD} $$ This square is send via $H$ into the square $$\begin{CD} F(c)=H(c,0) @>>> H(c,1)=G(c) \\ @V{F(f)=H(f,\text{id}_0)}VV @VV{H(f,\text{id}_1)=G(f)}V \\ F(c')=H(c',0) @>>> H(c',1)=G(c') \end{CD}$$ which is a naturality square.

So as you can see the images of the morphisms $(c,0) \to (c,1)$ via $H$ forms the components if natural transformations. This happens because $C \times 2$ has enough structure to build the above mentioned square.

Of course there is a deeper reason for why this works, namely that $2$ is the classifying object for the morphisms of a category, that means that there is a natural bijection between functors from $2$ in a category $C$ and morphisms of $C$. But that is a story for another day.

At this point I hope that the answer to

  1. How do we get a bijection between a natural transformation and functor $H$ can be easily guessed.

Feel free to ask for any additonal detail.


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