This is from pages 9 and 10 of Topology: A Categorical Approach by Bradley, Bryson and Terilla.
Excerpt:
I'm trying to understand this functor and verify that it satisfies the rules for a functor as given by the excerpt below.
Verification:
(i) An object $Y$ in $\mathsf{Set}$ gets sent to the object $X \times Y$ in $\mathsf{Set}$.
(ii) A morphism $f \colon Y \to Z$ in $\mathsf{Set}$ gets sent to the morphism $\text{id} \times f \colon X \times Y \to X \times Z$ in $\mathsf{Set}$.
(iii) If $g \colon A \to B$ and $h \colon B \to C$ are morphisms in $\mathsf{Set}$, then their composition is $hg \colon A \to C$. Applying the functor to $hg$ we get $\text{id} \times hg \colon X \times A \to X \times C$. I need to show that this is equal to the composition of the morphisms $\text{id} \times g \colon X \times A \to X \times B$ and $\text{id} \times h \colon X \times B \to X \times C$.
I am not sure how to do this. I think I'm being bamboozled by the notation. I need to show that $(\text{id} \times h) \circ (\text{id} \times g) = \text{id} \times hg$. Is there some identity that I need to use that I'm not aware of?
(iv) If we apply the functor to $\text{id}_X \colon X \to X$ we get $\text{id} \times \text{id}_X \colon X \times X \to X \times X$, which shows that $F \text{id}_X = \text{id}_{FX}$ as desired.
Questions:
How should I handle part (iii)?
Is this functor a specific example of a more general type of functor? I ask because the next example given in the book (shown below) looks very similar.
I appreciate any help.
Best Answer
You show that two functions are equal by showing that they take the same values at all points (formally this is called extensional equality). So, evaluate each side on points and see what you get. You'll end up proving something more general called the interchange law
$$(f_1 \times f_2) \circ (g_1 \times g_2) = (f_1 \circ g_1) \times (f_2 \circ g_2)$$
which amounts to asserting that cartesian product $X \times Y$ is a functor in each variable. The same law holds for the tensor product and the proof is very similar.
It's annoyingly difficult to even write down things like this using "1-dimensional notation." The interchange law is about a million times more intuitive when written "2-dimensionally," using string diagrams, where it asserts that the following string diagram is unambiguous (see this blog post for details):