# Does $\frac{\Pi_A\otimes I_B}{\text{Tr}((\Pi_A\otimes I_B)\rho_{AB})}\rho_{AB}=\rho_{AB}$ hold for a state $\rho_{AB}$ and projector $\Pi_A$?

For some projector $$\Pi_A$$ and state $$\rho_{AB}$$, let

$$\sigma_{AB} = \frac{\Pi_A\otimes I_B}{\text{Tr}((\Pi_A\otimes I_B)\rho_{AB})}\rho_{AB}$$

Is it the case that $$\sigma_B = \rho_B$$? It seems intuitively true since the projector is acting only on the $$A$$ system but I'm not sure how to prove this.

No, this is not the case. Consider the situation where $$\rho_{AB}=\frac12(|0\rangle\langle 0|\otimes |0\rangle\langle 0|+|1\rangle\langle 1|\otimes |1\rangle\langle 1|).$$ So, we have that $$\rho_B=\text{Tr}_A(\rho_{AB})=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)$$.
Now let $$\Pi_A=|0\rangle\langle 0|$$, which means that $$\sigma_{AB}=|00\rangle\langle 00|$$ and hence $$\sigma_B=|0\rangle\langle 0|$$. Clearly, $$\sigma_B$$ and $$\rho_B$$ are different.