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Now on the final line is what is confusing me. What I understand by what the author is saying is that the morphism $\sigma \in \text{Hom}_{C_A}(f_1, f_2)$ is exactly the morphism $\sigma \in \text{Hom}_C(Z_1, Z_2)$ such that $f_1 = f_2 \circ \sigma$.

It is clear to me why $\sigma$ exists (because $C$ is a category and every morphism in $C$ has a source and a target object). However it is not clear to my why $f_2 \circ \sigma = f_1$. In the definition of a category there exists a mapping called composition, written by $(f, g) \mapsto g \circ f$, so $f_2 \circ \sigma$ certainly exists, but nothing in the definition of the category $C$ states that we need to have $f_1 = f_2 \circ \sigma$.

Also note that in the definition of a category, the mapping called composition is not surjective. So since $f_1 \in \text{Hom}_C(Z_1, A)$, there need not exist a $f_2 \circ \sigma$ with $f_2 \in \text{Hom}_C(Z_2, A)$ and $\sigma \in \text{Hom}_C(Z_1, Z_2)$ such that $(\sigma, f_2) \mapsto f_2 \circ \sigma = f_1$.

So what exactly does the author mean?

Perturbative
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    For $\sigma\in \text{Hom}(Z_1,Z_2)$ it is not necessary that $f_2\sigma=f_1$. The morphisms in the slice category are those $\sigma\in \text{Hom}(Z_1,Z_2)$ which do satisfy $f_2\sigma=f_1$. – Angina Seng Dec 21 '17 at 07:42

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It is true that an arrow $\sigma : Z_1 \to Z_2$ need not satisfy $f_1 = f_2 \circ \sigma$. If it doesn't, then it is not a member of $\hom_{C_A}(f_1, f_2)$. If it does, then it is a member of $\hom_{C_A}(f_1, f_2)$.

It can indeed be the case that there aren't any $\sigma : Z_1 \to Z_2$ at all that satisfy $f_1 = f_2 \circ \sigma$. In this case, $\hom_{C_A}(f_1, f_2) = \varnothing$.

  • Ahh so the elements of $\text{Hom}_{C_A}(f_1, f_2)$ are the morphisms that make the commutative diagram, not the actual commutative diagram itself correct? – Perturbative Dec 21 '17 at 09:15
  • @Perturbative: There's not really any difference between doing it either way, the triangle is completely determined by its three sides and two are given. Different encodings will give isomorphic categories. If you prefer for the homsets to be disjoint, though, you'll need to make the homset contain triangles, or use some other device, such as the usual one of taking the morphisms to be triples $(f_1, \sigma, f_2)$. (to be explicitly clear, it is possible for $\sigma$ to witness elements of two different homsets; i.e. you can have $f_2 \circ \sigma = f_1$ and $g_2 \circ \sigma = g_1$) –  Dec 21 '17 at 09:24
  • I'm new to Category Theory, so I apologize if this is a silly question, but why would one prefer the homsets to be disjoint? – Perturbative Dec 21 '17 at 09:52
  • @Perturbative: Often people want to talk about the class of arrows in a category. If the homsets are disjoint, then this class can be expressed very simply as being the union of the homsets. AFAIK all of the reasons for this preference basically amount to this one. –  Dec 21 '17 at 09:55
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I will denote the slice category $\mathbf{C}$ over $A$ as $\mathbf{C}\downarrow A$.

You are looking at it from the wrong angle, I suppose. He is saying that if there exists a morphism $\sigma\in \mathrm{Hom}_\mathbf{C}(Z_1, Z_2)$ that makes the triangle commute ($f_1 = f_2\circ\sigma$), then we define a canonical morphism $\hat\sigma\in\mathrm{Hom}_{\mathbf{C}\downarrow A}(f_1, f_2)$, for $f_i:Z_i\rightarrow A$. In particular that homset in $\mathbf{C}\downarrow A$ could possibly be empty.

Phrased differently, he only defines morphism in the slice category via morphisms that make triangles commute - others are not taken into consideration, you don't need to prove anything.

Well, of course you still have to check that $\hat{\sigma}$ satisfies the definition of a morphism, but that is quite easy, and you should definitely do that as an exercise ;)

Jo Mo
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  • The elements of $\text{Hom}_{C_A}(f_1, f_2)$ are the morphisms that make the commutative diagram, not the actual commutative diagram itself correct? – Perturbative Dec 21 '17 at 09:15
  • Yes, precisely. And the commutativity of these diagrams is then what you would use to show that $\hat\sigma\circ\hat\tau = \widehat{\sigma\circ\tau}$ – Jo Mo Dec 21 '17 at 09:19