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 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 $g=h$.