I am currently reading the book "Introduction to modern cryptography" by katz et al. In this book there is a proof that looks something like this:
And written by myself it looks like this, even though the picture is way better:
\begin{align} Pr[C=c] &= \sum_{m\in \mathcal{M}} \Pr[C=c_j\mid M=m] \Pr[M=m] \\ &= \sum_{m\in \mathcal{M}} \gamma \cdot \Pr[M=m] \\ &= \gamma \cdot \sum_{m\in \mathcal{M}} \Pr[M=m] \\ &= \Pr[C = c|M = m]\\ \end{align}
What I don't understand is the line where we just move the gamma outside of the sigma notation? surely the value the summed value of all c's times gamma, is not the same as every value times gamma summed?
Is this allowed? to just move a value outside of sigma like this?
The second part of the proof is trying to show that if the conditional distribution of $C$ given $M$ is constant with respect to the outcome of $M$, then the probability $\Pr[C = c \mid M = m] = \Pr[C = c]$. We then call $C$ independent of $M$. This is why they are able to "move" the $\gamma$ term outside of the sum, because the term $\Pr[C = c \mid M = m]$ is the same $\gamma$ for each $m \in \mathcal M$, these terms can be factored out of the summation, leaving behind $$\sum_{m \in \mathcal M} \Pr[M = m] = 1.$$ It is like saying $$\sum_{k=1}^n 2k = 2 \sum_{k=1}^n k = 2 \cdot \frac{n(n+1)}{2} = n(n+1).$$ We just factor out the $2$ because it doesn't depend on the index of summation $k$. In the proof, it is by construction as one of the conditions of the proof that for all $m_0, m_1 \in \mathcal M$, any distribution $M$, every $c \in \mathcal C$, that $\Pr[C = c \mid M = m_0] = \Pr[C = c \mid M = m_1]$. This means for every $m \in \mathcal M$, $\Pr[C = c \mid M = m]$ equals some constant, which they call $\gamma$, that is the same for any choice of $m \in \mathcal M$. The ability to do this is certainly not true in general, such as when the probability of observing some outcome of $C$ does depend on the outcome of $M$. But the proof is addressing the specific case where it does not.