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In some categories, to verify that a map $f : X \to Y$ is a morphism, it suffices to check only a generating set for $Y$ (or, rather, a generating set for some structure on $Y$ such as a topology or a $\sigma$-algebra).

Here are two examples:

  • If $(X, \mathcal{T})$ and $(Y, \mathcal{T}')$ are topological spaces, in order to show that a function $f : X \to Y$ is continuous, it suffices to check that for every element $S$ of a basis or subbasis of $\mathcal{T}'$, we have $f^{-1}(S) \in \mathcal{T}$.

  • If $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ are measurable spaces, in order to show that $f: X \to Y$ is measurable, it suffices to check that for every element $B$ of a generating set for $\mathcal{B}$, we have $f^{-1}(B) \in \mathcal{A}$.

Is there any deeper meaning to this phenomenon, or does it simply arise "by nature"? More concretely, are there any nontrivial properties that a category must have in order for it to be true that in order to verify that a map is a morphism, it suffices to check a generating set for the codomain of the map?

As Zhen Lin noted, a category with this property must be concrete, but what else can be said about it?

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This question only makes sense for concrete categories, and very concrete ones at that: these generating sets are really just sets and not objects in the same category. – Zhen Lin Dec 11 '12 at 12:07
Some of them can be realised as morphisms though, e.g. a basis of $V$ can be thought of as an isomorphism $V\to\mathbb{F}^n$ in the category of vector spaces over $\mathbb{F}$ (or even any set of fixed reference vector spaces, one for each dimension/isomorphism class). However I agree that a lot of abstraction will be needed to characterize the other examples in this way, if it is possible at all. – Matthew Pressland Dec 11 '12 at 17:02
The claims you make for groups and for vector spaces are false. Consider the function $f : \mathbb{R}^3 \to \mathbb{R}$ given by $f(1,1,1) = 4$ and for all other vectors $f(x,y,z) = x+y+z$. This functions is not linear (and not a group homorphism) but it passes your criterion for the standard basis of $\mathbb{R}^3$. – Omar Antolín-Camarena Dec 11 '12 at 18:58
@Omar You're absolutely right. I'll edit my question to remove the erroneous examples. – Andrew Uzzell Dec 11 '12 at 21:49

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