Isomorphism between (even, odd) and (positive, negative) This is a problem in Conceptual Mathematics: A First Introduction to Categories

Find an isomorphism
$$
(\{{\rm odd}, {\rm even}\}, +) \stackrel{f}{\to} (\{{\rm positive}, {\rm negative} \}, \times)
$$
HINT: There are only two invertible maps of sets from $\{{\rm odd}, {\rm even}\}$ to $\{{\rm positive}, {\rm negative} \}$. One of them 'respects the combining rules', but the other doesn't

I've a little stumped with this question. This is one of the maps I've come up with
$$
f(x): \begin{cases}
-k &amp; {\rm if} &amp; x = 2k - 1, ~~k=1,2, ...\\
k &amp; {\rm if} &amp; x = 2k, ~~ k = 0, 1, 2,
\end{cases}
$$
This will map odd to positive numbers, and even to negative numbers, and is invertible. The issue is that it does not preserve the combining rules, e.g. if $x = 2m - 1$ is odd, and $y = 2n$ even
$$
f(x + y) = f(2(m + n) - 1) = -(m + n) \tag{1}
$$
but on the other hand
$$
f(x) \times f(y) = f(2m + 1) \times f(2n) = (-m) \times (n) = -mn \tag{2}
$$
Clearly
$$
f(x + y) \not= f(x)\times f(y)
$$
Any other suggestion?
Let g be the map of sets from {odd,even} to {pos, neg} that sends odd to pos and even to neg.
g is an isomorphism for obvious reason (existence of an inverse)
Now, suppose you want to &quot;add structure&quot; to g and make it an isomorphism between $(\{{\rm odd}, {\rm even}\}, +)$ and $(\{{\rm pos}, {\rm neg} \}, \times)
$. Let's prove that the following equation doesn't not hold:
$g(odd + even) \stackrel{?}{=}  g(odd) * g(even)$
$g(odd + even)    = g(odd)$  (odd + even = odd, combining rule for elements in {odd,even})
             = pos             g(odd) = pos by construction of g

g(odd) * g(even) = pos * neg       by construction of g
             = neg             (pos * neg = neg, combining rule in ({pos, neg}, *))

g(odd+even) is different from g(odd) * g(even) that's why the combining rule is not followed by g.
Let g be the map of sets from {odd,even} to {pos, neg} that sends odd to pos and even to neg.
g is an isomorphism for obvious reason (existence of an inverse)
Now, suppose you want to &quot;add structure&quot; to g and make it an isomorphism between $(\{{\rm odd}, {\rm even}\}, +)$ and $(\{{\rm pos}, {\rm neg} \}, \times)
$. Let's prove that the following equation doesn't not hold:
$g(odd + even) \stackrel{?}{=}  g(odd) * g(even)$
$g(odd + even)    = g(odd)$  (odd + even = odd, combining rule for elements in {odd,even})
             = pos             g(odd) = pos by construction of g

g(odd) * g(even) = pos * neg       by construction of g
             = neg             (pos * neg = neg, combining rule in ({pos, neg}, *))

g(odd+even) is different from g(odd) * g(even) that's why the combining rule is not followed by g.