Let $X$ be a normal variety over a field. Let $\Lambda$ be a nonempty linear system of divisors on $X$. The fixed part of $\Lambda$ is defined to be the effective divisor $$ \operatorname{Fix}\Lambda=\min_{D\in\Lambda}D, $$ where the minimum is taken in each component.
Let $Y$ be another normal variety and $f\colon Y\to X$ be a morphism. The pullback of $\Lambda$ is the linear system on $Y$ $$ f^*\Lambda = \{f^* D\mid D\in\Lambda\}. $$
Question. Is it true that $$ \operatorname{Fix} f^*\Lambda = f^*(\operatorname{Fix}\Lambda)? $$
My attempt: I proved that $\operatorname{Fix} f^*\Lambda \ge f^*(\operatorname{Fix}\Lambda).$ Indeed, for any $D\in\Lambda$, we have $f^*D\ge f^*(\operatorname{Fix}\Lambda)$ since $D\ge \operatorname{Fix}\Lambda$.
How can we show the convese? Is there a counter exmple? I cannot proceed here. Thanks in advance.