Reading through Joy of cats. I am stuck at the reflective sub-categories. I can't understand how a reflective sub-category be not full (I couldn't understand the last two examples of the section for the same reason). Suppose, a function $f: A \rightarrow A' \notin Hom_X(A, A')$ where $X$ is a subcategory of $Y$ and $A, A' \in X$ then for $B \in Y$ every morphism $g: B \rightarrow A'$ has to have a reflection, lets call it $h$ now $f \circ h \in Y$ but if $f \notin X$ $A$ would't be a reflection now right?

  • $\begingroup$ "then for B∈Y every morphism g has to have a reflection" - What are the domain and codomain of $g$ you have in mind here? What's $g$'s relation to $B$? $\endgroup$ May 6, 2020 at 7:17
  • $\begingroup$ I edited. Can you please check if it makes sense? $\endgroup$ May 6, 2020 at 9:12

1 Answer 1


Let me try to understand what you said with your $X,Y,A,A',B$, and then I'll give an example to see what's happening.

You take $B\in Y$, and a reflection of it in $X$, say $h: B\to A$. If $f: A\to A'$ is not in $X$, you get $f\circ h : B\to A'$. You say "this must factor through some $A\to A'$ which is in $X$; and since you chose $f$ not to be in $X$, this seems impossible.

But the point is that there will be some $f' : A\to A'$ such that $f'\circ h = f\circ h$, with $f'\in X$. And this $f'$ will be unique in $X$.

An easy example of a non full reflective subcategory is, given any category $C$ with products, the diagonal $\Delta: C\to C^2$.

We may clearly see it as a subcategory: the subcategory on objets $(A,A)$ and morphisms $(f,f)$ between those. In general, it's not full : there will, in general, be objects $A,B$ with two distinct morphisms $f,g:A\to B$ (if $C$ is not a poset, you're guaranteed to find such $A,B$), and so $(f,g) : (A,A)\to (B,B)$ is not in the image of $C$.

A left adjoint to this inclusion is given by the coproduct $(A,B)\mapsto A\coprod B$ (and if you want to really view it as a subcategory, it's $(A,B)\mapsto (A\coprod B, A\coprod B)$)

It's clear that this is not a full reflective subcategory, as if it were, then the reflector applied to $(A,A)$ would be just $(A,A)$ : here it's $(A\coprod A, A\coprod A)$ which is in general wildly different.

Now in my example, what's happening with your question : take $f,g: A\to B$ that are different, so we get $(f,g) : (A,A)\to (B,B)$ which is not in $\Delta(C)$, and suppose $(A,A)$ is the reflection of some $(E,F)$, that is $A= E\coprod F$.

Then $f,g$ are determined by $f_0,g_0: E\to B$ and $f_1,g_1: F\to B$, and the map $(E,F)\to (B,B)$ is given by $(f_0,g_1)$. But then its reflection is $([f_0,g_1],[f_0,g_1]) : (A,A)\to (B,B)$, which is different from what we started with, i.e. $(f,g) = ([f_0,f_1],[g_0,g_1])$ .

So we see that we do get a different f$'$ which will be in $X$, and which will be the only map in $X$ to satisfy $f'\circ h = f\circ h$.

(where for maps $h : E\to D, k:F\to D$, I let $[h,k]: E\coprod F\to D$ denote the uniquely determiend map)

  • $\begingroup$ I think I understood the product example. Though I got lost on left adjoint (Not sure what it is yet). But I see that product subcategory with just diagonals as morphisms definitely has a reflection. I also have another question, if the factored function as you said is not unique (suppose both $f$ and $f'$ is in the subcategory) is that a reflective test or it can never happen? $\endgroup$ May 7, 2020 at 17:13
  • $\begingroup$ Well if both $f,f'$ are in the subcategory, then the subcategory is not reflective, or the map $h: B\to A$ is not a reflection $\endgroup$ May 7, 2020 at 17:25
  • $\begingroup$ One last question, please, why is $id_X$ only a reflection of $X$ (Since its in $B$) only if its a full subcategory? $\endgroup$ May 7, 2020 at 17:29
  • $\begingroup$ This is proved in Joy of cats if I recall correctly $\endgroup$ May 7, 2020 at 17:39
  • $\begingroup$ No it just hand waves it as obvious :( $\endgroup$ May 7, 2020 at 18:17

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