# How can a reflective subcategory be non full?

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?

• "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 $B$? May 6, 2020 at 7:17
• I edited. Can you please check if it makes sense? May 6, 2020 at 9:12

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)

• 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? May 7, 2020 at 17:13
• 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 May 7, 2020 at 17:25
• One last question, please, why is $id_X$ only a reflection of $X$ (Since its in $B$) only if its a full subcategory? May 7, 2020 at 17:29
• This is proved in Joy of cats if I recall correctly May 7, 2020 at 17:39
• No it just hand waves it as obvious :( May 7, 2020 at 18:17