What is the identity arrow in this category? This is taken from An Introduction to Category Theory by Harold Simmons (Example 1.3.1) (slightly changed for formatting reasons).

The objects are the finite sets.  An arrow $f$ with source $A$ and target $B$ is a function $$f : A\times B \to \mathbb{R}$$ with no imposed conditions.  For each pair of arrows $f,g$ with $\text{source}(f) = A$, $\text{target}(f) = B$, $\text{source}(g) = B$, $\text{target}(g) = C$ we define $g \circ f$ to be a function $$g \circ f : A \times C \to \mathbb{R}$$ with $$(g\circ f)(a,c) = \sum_{y \in B} f(a,y)g(y,c).$$

I understand how to check that for three arrows composition is associative, but I am unsure how one would define, for example, $\text{Id}_A, \text{Id}_B$ such that $$\text{Id}_B \circ f = f = f \circ \text{Id}_A,$$
I know they would be functions from $A\times A \to \mathbb R$ and $B\times B \to \mathbb R$ but when writing down the composition I am not sure how one obtains the original $f$.
 A: Writing down the relation that the identity $I_A$ would have to satisfy on one side : $$f(a,a') = (f \circ I_A)(a,a') = \sum_{a'' \in A} f(a,a'')I_A(a'',a'),$$
so a natural candidate would be $I_A(a'',a') = 1$ if $a'' = a'$ and $0$ otherwise. (This is known as the Kronecker delta function.)
A: $\newcommand{\id}{\mathsf{id}}$
You'd like a function $f: A\times A \rightarrow \mathbb{R}$ such that for any other function $g: A\times B \rightarrow \mathbb{R}$, you have
$$\begin{eqnarray*}
f \circ g &= g & \qquad A\times B\rightarrow \mathbb{R}\\
(f \circ g)(a,b) &= g(a,b) &\qquad \forall \,(a,b) \in A\times B\\
\sum_{y \in A} f(a, y)\cdot g(y, b) &= g(a,b) & \\
\end{eqnarray*}$$
We can define $f$ in such a way that most terms in the sum vanish. In particular, if we set $f(a,y) \equiv 0$ unless $y=a$, then the sum collapses to a single term:
$$f(a,a)\cdot g(a,b) = g(a,b)$$
Then if we define $f(a,a) = 1$, we get the desired result: $g(a,b) = g(a,b)$, so $f\circ g = g$ in general.
The definition we've arrived at is:
$$f(a,y) \equiv \begin{cases}0 & \text{if }a \neq y\\1 & \text{if }a=y\end{cases}$$
which you can prove is also a right handed identity: $h\circ f = h$.
A: You are looking for the Kronecker delta $\delta(x,y)$ which is $1$ if $x =y $ and $0$ otherwise. 
