universal property of product: must any map satisfying it be a morphism

I am thinking about the universal property of products:

Let $X$ and $Y$ be objects of a category $D$. The product of $X$ and $Y$ is an object $X \times Y$ together with two morphisms

$\pi_1 : X \times Y \rightarrow X$

$\pi_2: X \times Y \rightarrow Y$

such that for any other object $Z$ of $D$ and morphisms $f : Z \rightarrow X$ and $g : Z \rightarrow Y$ there exists a unique morphism $h: Z \rightarrow X \times Y$ such that $f = \pi_1 \circ h$ and $g = \pi_2 \circ h$.

Does it mean that of all morphisms, only one satisfies this, but there could be a map which is not a morphims which makes this work?

That is, if any map $h$ makes this diagram commute then it must necessarily be a morphism?

Sorry, if this is obvious - still getting my head around making sense of caterory theoretic formulations.

To explain where I am coming from: my projections and map $f,g$ are morphisms in category I am working in (say $A$, and I know finite products exists in $A$), I have found a $h$ in conceivably a different category (say $B$) which makes everything commute. I am trying to see if it's possible to argue by virtue of everything else being in $A$, that $h$ must also be a morphism in $A$. There's probably a silly counter example to this that I am missing.

• In category theory there's no such thing as a map which is not a morphism. – Qiaochu Yuan Feb 10 '16 at 5:01

In the context of category theory, all "maps" are morphisms. There are no other "maps". If, however, you happen to know that the morphisms in your category correspond to morphisms in another category (typically via a functor), then you can talk about whether a product in one category is a product in the corresponding category. Put a little more formally, if $U : \mathcal{C} \to \mathcal{D}$ is a functor which takes morphisms in your category to morphisms in another category (and we'll call morphisms in this other category "maps"), your question is: if $p_1: UX \to UA$ and $p_2 : UX \to UB$ form a product in $\mathcal{D}$, then is it necessarily the case that $p_1 = U\pi_1$ and $p_2 = U\pi_2$ for unique $\pi_1$ and $\pi_2$ which form a product in $\mathcal{C}$? If this is the case, we say $U$ creates products (or, for general limits, $U$ creates limits).

Taking the example from Wikipedia, if $U : \mathbf{Grp} \to \mathbf{Set}$ where $U$ is the underlying set functor which takes a group and simply forgets the extra structure, then $U$ creates limits. Then from your example or the above, you can think of $h$ as a group homomorphism (a morphism of $\mathbf{Grp}$) and $Uh$ as a set function or "map" (a morphism of $\mathbf{Set}$). Then, for this example, the property you talk about holds. Creating limits is not at all guaranteed, though it is an attribute of algebraic cases. For a counter-example, if $U : \mathbf{Top} \to \mathbf{Set}$ then the morphisms of $\mathbf{Top}$ are continuous functions, where the product in the underlying category is essentially what we'd get for the discrete topology, but if we had another topology...

• yes, this was sort of what I was getting at. still need to think about this and see if I can make it work. thanks. – sgldiv Feb 10 '16 at 5:38

"Does it mean that of all morphisms, only one satisfies this, but there could be a map which is not a morphims which makes this work?"

Why are you making a distinction between maps and morphisms? By the definition of a category, there is a set of morphisms $Hom_D(Z,X\times Y)$. The condition that $X\times Y$ is the product of $X$ and $Y$ is just saying that for every object $Z$ in $D$, and morphisms $f:Z\to X$, $g:Z\to Y$, the set $$\{h\in Hom_D(Z,X\times Y) \ | \ f=\pi_1 h,g=\pi_2 h\}$$ has only one element.

• I think I am trying to guard against the case where say $X,Y$ are groups, $h$ could be morphism when considering them as sets, but $h$ does not respect the group structure so is not a morphism in category of groups?? Does that make sense? – sgldiv Feb 10 '16 at 5:09
• @sgldiv: in that case you're no longer working in the category of groups; you're implicitly passing to the category of sets. – Qiaochu Yuan Feb 10 '16 at 5:15
• Well, when you're working in a category, you don't have to care about what's going on in other categories. So if you're working in the category of groups, you don't have to care about morphisms that aren't group homomorphisms. Now, if you do want to care, for most algebraic categories, there is a forgetful functor to Sets and this functor is right adjoint. So it will preserve products, and therefore there is also a unique mophism of set to the product (which happens to be a morphism of group, module, ring,...). But in general, there isn't a forgetful functor. – Nitrogen Feb 10 '16 at 5:15
• To explain where I am coming from: my projections and map $f,g$ are morphisms in category I am working (say $A$), I have found a $h$ in conceivably a different category (say $B$). I am trying to see if it's possible to argue by virtue of everything else being in $A$, that $h$ must also be a morphism in $A$. There's probably a silly counter example to this that I am missing. – sgldiv Feb 10 '16 at 5:23
• You could do something like that, but you need a way to relate $A$ and $B$ by a functor if you want to compare them in that way. – Nitrogen Feb 10 '16 at 5:34

Not all categories are of the form "objects = sets with structure", and "morphism = structure preserving map/function between such sets". In general category theory this question makes no sense, cannot even be asked. There are only "morphisms" (or arrows) between objects and these objects need not be sets, nor need the arrows/morphisms be functions.

Of course, we can ask such questions in a wider context (e.g. in categorical topology, where we do have this extra convention of structure preserving sets, and we should have initial and final structures). For topological spaces, e.g. what you're asking is in fact true, as it is for groups and rings etc. as far as I can tell. If we start with morphisms (which in that case are special functions), the unique function we get (from the fact this morphism = function exists in $\textbf{Set}$) is in fact a morphism in that category. But this works in settings like Categorical Topology or Universal Algebra, which are less broad than category proper.