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 ?

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    $\begingroup$ 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? $\endgroup$
    – lisyarus
    Jan 24, 2021 at 15:21
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    $\begingroup$ 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. $\endgroup$ Jan 24, 2021 at 15:22
  • $\begingroup$ @lisyarus Thanks, yes, and any further restrictions necessary for the proof. $\endgroup$ Jan 24, 2021 at 15:30
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    $\begingroup$ @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) $\endgroup$
    – qualcuno
    Jan 24, 2021 at 15:32

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Jan 25, 2021 at 14:12
  • $\begingroup$ 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 $\endgroup$
    – qualcuno
    Jan 25, 2021 at 21:45

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