Show that if $\langle f,g\rangle=\langle k,h\rangle$ then $f=k$ and $g=h$

I attempted this but I am feeling very uncomfortable because I don't think the things I assumed about $\langle f,g\rangle=\langle k,h\rangle$ are right. Would anyone mind checking my proof, and perhaps commenting on my questions below please?

In particular, just because we have a pair of arrows does not necessarily mean that the product is involved...right?

In addition, I seem to have trouble understanding the concept of product in categories on the whole. Is $a \times b$ supposed to be a Cartesian product? But isn't that a set theoretic concept?

In the book, it was also mentioned that category theory deals with Cartesian product without mentioning ordered pair; but a pair of arrows seems almost indistinguishable from an ordered pair to me.

Def: A product in a category of two objects $a$ and $b$ is an object $a \times b$ together with a pair $(pr_a: a\times b \to a$,$pr_b: a\times b \to b)$ of arrows such that for any pair of arrows of the form $(f:c\to a, g: c\to b)$ there is exactly one arrow $\langle f, g \rangle : c \to a \times b$ making enter image description here commute, ie. such that $pr_a \circ \langle f, g \rangle =f$ and $pr_b \circ \langle f, g \rangle =g$ is the product arrow of $f$ and $g$ with respect to the projections $pr_a$, $pr_b$.

Here I will make a wild guess - that $\langle f, g\rangle$ and $\langle k, h\rangle$ are the arrows from one object (let's call it $D$) going into a Cartesian product $a \times b$ (ie. they are like the $\langle f, g\rangle$ as shown in the def. image).

If so, then this means $pr_a \circ \langle f, g \rangle =f$ and $pr_a \circ \langle k, h \rangle =k$. But because $\langle f,g\rangle=\langle k,h\rangle$, this means $pr_a \circ \langle k, h \rangle =f$.

However, $pr_a$, being a projection function, separates out the left coordinate. If given $\langle k, h \rangle$ $pr_a$ separates out $f$ instead, this should mean that $f=k$. And a similar reasoning goes with $g=h$.

  • 1
    $\begingroup$ Yes, $\langle f,g\rangle$ here doesn't denote an ordered pair, but the unique arrow $c\to a\times b$ that makes the diagram commutative. $\endgroup$
    – Berci
    Oct 3, 2019 at 16:53

2 Answers 2


The notation $a \times b$ can be confusing when you're first learning about products in categorical terms. $a \times b$ is simply an object that happens to satisfy the universal property of the product. Depending on the underlying category, it may or may not be the cartesian product. Similarly, $\langle f, g\rangle$ is the arrow induced by the universal property. It's certainly not an ordered pair.

To illustrate, we can rewrite the definition of the product without using any suggestive notation. The product of two objects $a$ and $b$ in a category $\mathcal C$ is an object $d$ with a pair of arrows $p_a : d \to a$ and $p_b : d \to b$ such that for any object $c$ and arrows $f : c \to a$ and $g : c \to b$, there exists a unique arrow $u : c \to d$ satisfying $f = p_a \circ u$ and $g = p_b \circ u$.

Your proof of the question given is exactly right. Since $\langle f, g\rangle = \langle k, h\rangle$, we have $$ f = p_a \circ \langle f, g\rangle = p_a \circ \langle k, h\rangle = k. $$

And similarly, $g = h$.

  • $\begingroup$ Thank you, not sure what universal property (it seems to deal with functors, which is in later chapters of my book) is, but your 2nd paragraph made things clearer from a different perspective. $\endgroup$ Oct 3, 2019 at 20:48

Briefly, you deduced correctly the equalities $pr_a \circ \langle k,h\rangle=k$ and $pr_a \circ \langle k,h \rangle =f$, right? Hence $k=pr_a \circ \langle k,h \rangle =f$. And as you said the same works for $pr_b$.

Observe that a stronger property holds: $\langle f,g \rangle =\langle k,h \rangle$ if and only if $f=k$ and $g=h$. This is precisely the definition of product you wrote.


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