Why correlation property of $\mathcal Z$-transform contains a time reversal operation I'm reading through Digital Signal Processing, Proakis and Manolakis, third edition.  I've reached section 3.2: Properties of $\mathcal Z$-transform.

One property is the convolution:

$$x(n) = x_1(n)\star x_2(n) \longleftrightarrow X(z) = X_1(z)X_2(z)$$

Other property is correlation:

$$r_{x_1,\ x_2}(l) = \sum_{n=-\infty}^{\infty}x_1(n)x_2(n-l) \longleftrightarrow R_{x_1,\ x_2}(z)=X_1(z)X_2\left(z^{-1}\right)$$

I'm confused by the time reversal (folding) which appears in the correlation property at $X_2\left(z^{-1}\right)$. I know that the folding step occurs on convolution but not on cross/auto/correlation. Why is this and what am I missing?

To clarify the "time reversal" expression I'm referring to another property of $\mathcal Z$-transform taken from the same source:

If $$x(n) \longleftrightarrow X(z)$$
Then
$$x(-n) \longleftrightarrow X(z^{-1})$$

For the two answers: Matt's one clarifies the matter using a mathematical perspective while Peter's one shows in an intuitive way how the folding occures in polynomial multiplication. So thanks, I wish SE had a way to mark two answers as accepted.
Note that (discrete-time) convolution is defined as

$$x_1[n]\star x_2[n]=\sum_kx_1[k]x_2[n-k]\tag{1}$$

and correlation is defined as

$$r_{x_1,x_2}[n]=\sum_kx_1[k]x_2[k-n]\tag{2}$$

Comparing $(1)$ and $(2)$ we see that correlation can be written as the following convolution:

$$r_{x_1,x_2}[n]=x_1[n]\star x_2[-n]\tag{3}$$

The $\mathcal{Z}$-transform of $x_2[-n]$ is given by

$$\sum_nx_2[-n]z^{-n}=\sum_nx_2[n]z^n=X_2\left(\frac{1}{z}\right)\tag{4}$$

Consequently, from $(3)$ the $\mathcal{Z}$-transform of $r_{x_1,x_2}[n]$ must be

$$R_{x_1,x_2}(z)= X_1(z)X_2\left(\frac{1}{z}\right)\tag{5}$$