# Category Theory: Equality if All Projections are Equal?

Let $$\{C_\alpha\}_{\alpha \in A}$$ be objects in a category: in response to comments, a concrete category.

Let the product object $$C = \Pi C_{\alpha \in A}$$ exist with corresponding projections {$$\pi_\alpha$$}

If $$x, y \in C$$ and $$\pi_\alpha (x) = \pi_\alpha (y)$$ for all $$\alpha$$ does it follow that $$x = y$$ ?

I think I can prove this in the category of vector spaces by assuming the form of $$x = (x_\alpha)$$ and linearity of the projections, but is it readily provable (even true ?) in general, say from the universal property ?

• What do you mean by $x,y \in C$? Objects in categories don't have a notion of "elements" in general. Are you working with a concrete category? Jan 24, 2021 at 15:21
• In a general category it doesn't really make sense to speak of elements of an object. If your category has a final object $X$ you can think of "morphism from $X$ to $C$" as a generalization of "object of $C$," but this doesn't always behave as you'd expect. Jan 24, 2021 at 15:22
• @lisyarus Thanks, yes, and any further restrictions necessary for the proof. Jan 24, 2021 at 15:30
• @lisyarus iirc it is standard to note $x \in \mathsf C$ to mean $x \in \mathsf{ob \ C}$, e.g. Riehl's book Category Theory in Context does this (edit: nvm, $C$ was an object, I stand corrected) Jan 24, 2021 at 15:32

You category is concrete, i.e. you have a fully faithful functor $$U \colon \mathsf{C} \to \mathbf{Set}$$. Suppose this has some right adjoint $$F \colon \mathbf{Set} \to \mathsf{C}$$ i.e. a "free construction".

Elements of $$C$$ as a set are maps $$x,y \colon \ast \to UC$$ in the category of sets. This is the first step - but very much in the spirit - in the direction of generalized elements. You want to show $$x = y$$, as elements of the set $$UC$$, which is the same as saying that they are the same arrow in $$\mathbf{Set}$$.

By the adjunction, this is to say that the maps $$x^\flat,y^\flat \colon R(\ast) \to C$$ induced by the adjunction coincide.

Since $$R$$ is a right-adjoint, it preserves limits and sends $$\ast$$ to the terminal object $$1$$ in $$\mathsf{C}$$, so we ought to decide wether $$x^\flat,y^\flat \colon 1 \to C$$ are equal, and since $$C$$ is a product, it suffices to check wether composing with each projection they do, that is wether $$\pi_\alpha x^\flat = \pi_\alpha y^\flat$$ for all $$\alpha$$.

But these are maps $$1 \to C_\alpha$$ which correspond via $$U$$ with $$\ast \to UC_\alpha$$ associated with the elements $$\pi_\alpha(x)$$ and $$\pi_\alpha(y)$$, assumed to be equal, completing the proof. Here we use that $$U 1 = \ast$$, as left adjoints preserve finite limits.

Just to be clear: the hypothesis I introduced were

• the elements $$x,y$$ are maps $$x,y \colon \ast \to UC$$.
• the forgetful functor $$U$$ has some right adjoint.
• for each projection we have $$U\pi_\alpha x^\flat = U \pi_\alpha y^\flat$$.

This coincides with the intuitive notion for classical concrete categories such as the category of vector spaces, topological spaces, abelian groups, etc.

Edit: the adjunction hypothesis is only to have a well defined notion of starting with an element of a set (i.e a map $$x : \ast \to UC$$) and lifting it to a map $$x^\flat \colon 1 \to C$$ in $$\mathsf{C}$$.

In general, what is true is that maps $$f,g \colon B \to C$$ are equal iff $$\pi_\alpha f = \pi_\alpha g$$ for all $$\alpha$$, but since $$U$$ is fully faithful this occurs precisely if $$U \pi_\alpha f = U \pi_\alpha g$$ for all $$\alpha$$. Hence we can check at the level of functions.

So if "elements" for you are maps $$1 \to C$$ to begin with, you only need the concreteness hypothesis.

• If "elements" of $C$ are arrows with codomain $C$, then you don't need concreteness - the statement follows immediately from the universal property of the product. Jan 25, 2021 at 14:12
• If your hypothesis is $U\pi_\alpha f = U \pi_\alpha g$, you still need $U$ to be fully faithful, although it doesn't need to be set-valued. If that's not our hypothesis then I think the question has lost pretty much all meaning; it would just be "do products exist?". My point was that you can still have a result in the positive direction even if you don't start with maps in the category of sets; you can still check equality function-wise. I prefer the first interpretation though, where we have an adjoint Jan 25, 2021 at 21:45