This is the first exercise in the section that introduces Categories in Aluffi's Algebra text (Ch.1 Sec.3). Since this is my first exposure to categories, I wanted to know if:
- I'm properly understanding the basic definition of a category
- My logic is correct
- The language and notation in my proof are idiomatic (e.g. Aluffi seems to use set-theoretic notation like $\in$ in the exposition).
Exercise 3.1:
Let $\mathsf{C}$ be a category. Consider a structure $\mathsf{C}^{op}$ with
- $\mathrm{Obj}(\mathsf{C}^{op}) := \mathrm{Obj}(\mathsf{C})$
- for $A,B$ objects of $\mathsf{C}^{op}$ (hence objects of $\mathsf{C}$), $\mathrm{Hom}_{\mathsf{C}^{op}}(A,B) := \mathrm{Hom}_{\mathsf{C}}(B,A)$.
Show how to make this into a category (that is, define composition of morphisms in $\mathsf{C}^{op}$ and verify the properties listed in $\S3.1$).
Intuitively, the 'opposite' category $\mathsf{C}^{op}$ is simply obtained by 'reversing all the arrows' in $\mathsf{C}$.
My Solution (original; see below for updated solution based on feedback):
Remember that by definition, a category must have i) an identity morphism for all objects in the category and ii) a composition morphism for any pairs of morphisms. Additionally, the composition must satisfy two additional properties: a) associativity and b) unital (e.g. for $f: X \to Y, f1_X = 1_Yf = f$).
i) Identity morphism: If $A$ is an object in $\mathsf{C}^{op}$, it also exists in $\mathsf{C}$ (by definition of $\mathrm{Obj}(\mathsf{C}^{op})$). Since $\mathsf{C}$ is a category, it satisfies the property of having an identity morphism for each object. Since we chose $A$ in $\mathsf{C}^{op}$ arbitrarily, it follows that every object in $\mathsf{C}^{op}$ has an identity morphism.
ii) Composition morphism: Let's define composition of morphisms as follows: for $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B)$ and $g \in \mathrm{Hom}_{\mathsf{C}^{op}}(B,C)$, define $fg \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,C)$ such that $fg = gf \in \mathrm{Hom}_{\mathsf{C}}(C,A)$. We know $gf$ exists because $\mathsf{C}$ is a category and thus satisfies the condition of having a composition morphism for all pairs of morphisms.
To prove that the composition morphism is both associative and unital:
a) Associativity: Let $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B), g \in \mathrm{Hom}_{\mathsf{C}^{op}}(B,C),$ and $h \in \mathrm{Hom}_{\mathsf{C}^{op}}(C,D)$. Then $(fg)h = (gf)h = h(gf) = (hg)f = (gh)f = f(gh)$. Where the 1st, 2nd, 4th, and 5th equalities are due to the definition we chose for the composition of morphisms, and the 3rd equality is true because $\mathsf{C}$ is a category and thus is associative.
b) Unital: Let $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B)$, then:
- $f1_A = 1_Af = f$
- $1_Bf = f1_B = f$
Where for both statements, the first equality is due to our definition of composition and the second equality is true because $\mathsf{C}$ is a category and thus is unital.
Thank you
Solution (updated):
Remember that by definition, a category must have i) a composition morphism that satisfies associativity for any pairs of morphisms and ii) an identity morphism that is unital for all objects in the category (e.g. for $f: X \to Y, f1_X = 1_Yf = f$).
i) Composition morphism: Let's define composition of morphisms as follows: for $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B)$ and $g \in \mathrm{Hom}_{\mathsf{C}^{op}}(B,C)$, define $f \circ{'} g \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,C)$ such that $f \circ{'} g = g \circ f \in \mathrm{Hom}_{\mathsf{C}}(C,A)$. We know $g \circ f$ exists because $\mathsf{C}$ is a category and thus satisfies the condition of having a composition morphism for all pairs of morphisms.
To show that the composition morphism is associative, let $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B), g \in \mathrm{Hom}_{\mathsf{C}^{op}}(B,C),$ and $h \in \mathrm{Hom}_{\mathsf{C}^{op}}(C,D)$. Then $(f \circ{'} g) \circ{'} h = (g \circ f) \circ{'} h = h \circ (g \circ f) = (h \circ g) \circ f = (g \circ{'} h) \circ f = f \circ{'} (g \circ{'} h)$. Where the 1st, 2nd, 4th, and 5th equalities are due to the definition we chose for the composition of morphisms, and the 3rd equality is true because $\mathsf{C}$ is a category and thus its morphisms are associative.
ii) Identity morphism: If $A$ is an object in $\mathsf{C}^{op}$, it also exists in $\mathsf{C}$ (by definition of $\mathrm{Obj}(\mathsf{C}^{op})$). Since $\mathsf{C}$ is a category, it satisfies the property of having an identity morphism for each object. Let's define $1_A$ for any $A$ in $\mathsf{C}^{op}$ to be the same as $1_A$ for the same $A$ in $\mathsf{C}$.
To show that the identity morphism is unital, let $f \in \mathrm{Hom}_{\mathsf{C}^{op}}(A,B)$, then:
- $f \circ{'} 1_A = 1_A \circ f = f$
- $1_B \circ{'} f = f \circ 1_B = f$
Where for both statements, the first equality is due to our definition of composition and the second equality is true because $\mathsf{C}$ is a category and thus its identity morphisms are unital.
Best Answer
Seems generally correct, but you could be a bit more careful.
i) An 'identity morphism' is just a distinguished element of $Hom_\mathsf{C}(A,A)$. Given a choice of $1_A$ in the category $\mathsf{C}$, it makes sense to define the identity morphism of $A$ in $\mathsf{C}^{op}$ by choosing the same $1_A$, since you know $Hom_{\mathsf{C}^{op}}(A,A) = Hom_\mathsf{C}(A,A)$.
A morphism being the identity is not an intrinsic property of some particular function, it is a structural property of a morphism in a category, so if you are building a new category you need to say what the identities are.
ii)b Take a morphism $f\in Hom_{\mathsf{C}^{op}}(A,B)$ corresponding to $f'\in Hom_{\mathsf{C}}(B,A)$. The morphism $f\circ 1_A \in Hom_{\mathsf{C}^{op}}(A,B)$ is, by your definition of composition, the morphism corresponding to $1_A\circ f'\in Hom_{\mathsf{C}}(B,A)$ which is $f'\in Hom_{\mathsf{C}}(B,A)$. Thus $f\circ 1_A = f$.
You could just use the same letters for $f$ and $f'$, of course, I just wanted to emphasize that they 'go in different directions', i.e. 'live in different categories'.
Logically speaking I would prefer you write ii) and ii)a together and put ii)b and i) together: the composition is a law for combining morphisms, calling something an identity only makes sense after defining composition. You cannot call $0$ an additive identity without knowing what addition is.