Let $A$ be a unital $C^*$-algebra and $p,q \in A$ projections in $A$ with the property $$pqp = p, \quad qpq = q.$$ I am trying to show that $p=q$, but I don't really see why this should be the case. The obvious algebraic manipulations don't seem to work. Of course, we can WLOG assume that $A= B(H)$.
This question occurs because I'm trying to justify a step in the proof of Takesaki's first volume (chapter III, theorem 4.2, p141 below, where they deduce that $q= q_1$ from the equalities $q= qq_1q$ and $q_1 = q_1qq_1$).
Note that $0=p-pqp=p(1-q)p=(p(1-q))\cdot(p(1-q))^*$ and therefore by the $C^*$-identity we have $p(1-q)=0$, so $p=pq$. This is equivalent to $p\le q$. Likewise, since $0=q-qpq=q(1-p)q$, we get $q(1-p)=0$, qo $q=qp$ and thus $q\le p$. These two combined give $p=q$.