Say the index is
$$
X = \sum_{i=1}^n w_iX_i
$$
and the variance is ${\rm Var}(X) = \sigma^2$ then obviously the covariance of the index with itself is $\sigma^2$ so
$$
{\rm Cov}\left(\sum_{i=1}^n w_i X_i, X\right) = \sum_{i=1}^n w_i{\rm Cov}(X_i,X) = \sigma^2
$$
The correlation is the covariance divided by the square root of the produce of the variances, so dividing everything through by $\sigma^2$ you have
$$
\sum_{i=1}^n w_i \frac{{\rm Cov}(X_i,X)}{\sigma^2} = 1
$$
You might choose to consider the components of this sum the "contribution to the correlation" from each member of the index, i.e.
$$
\xi_i = \frac{w_i}{\sigma^2}{\rm Cov}(X_i,X)
$$
Noting that the covariance can be expressed in terms of the correlation, you also have
$$
\xi_i = \frac{w_i\sigma_i}{\sigma} {\rm Corr}(X_i, X)
$$
and finally, you could note that this is $w_i$ times the coefficient of regression of $X_i$ on $X$, commonly known as the beta, so you have
$$
\xi_i = w_i\beta_i
$$
It then becomes obvious that the $\xi_i$ must sum to one, since the average beta of the constituents to an index must be one.