Let $A$ be a $C^*$-algebra and $p,q \in A$ projections. We say that $p$ and $q$ are Murray - von Neumann equivalent, written $p \sim q$, if there is $u \in A$ with $p = uu^*$ and $q=u^*u$. I'm trying to show that $$p \sim q, \quad q \sim r \implies p \sim r$$
There are $u,v \in A$ with $$p = uu^*, q = u^* u = vv^* , r = v^* v$$
I want to find/construct $w \in A$ with $$p = ww^*, r = w^*w$$
How can I define $w$?
I feel like this should be simple. I must be missing something.
$u$ in the definition of Murray-von Neumann equivalence can't just be some arbitrary element of $A$; there are conditions imposed on it by the condition that $p$ and $q$ are projections. We have
$$p^2 = u u^{\ast} u u^{\ast} = p = u u^{\ast} = uqu^{\ast}$$ $$q^2 = u^{\ast} u u^{\ast} u = q = u^{\ast} u = u^{\ast} pu$$
Now we can set $w = uv$ (which doesn't obviously work!), which gives
$$ww^{\ast} = uv v^{\ast} u^{\ast} = uqu^{\ast} = uu^{\ast} = p$$ $$w^{\ast}w = v^{\ast} u^{\ast} u v = v^{\ast} qv = v^{\ast}v = r$$
as desired.
This is maybe still a bit too obscure so here is what is going on geometrically in the case $A = M_n(\mathbb{C})$. Then you can check that the condition that $P = X X^{\ast}$ for $P$ a projection matrix and $X$ a matrix means that the singular values of $X$ are equal to either $1$ or $0$ and that for the nonzero singular values $\sigma_1 = \dots = \sigma_k = 1$, the corresponding left singular vectors $u_1, \dots u_k$ of $X$ have the property that $\text{span}(u_1, \dots u_k) = \text{im}(P)$. Similarly the condition that $Q = X^{\ast} X$ says that the singular values of $X$ are equal to either $1$ or $0$ and the right singular vectors $v_1, \dots v_k$ have the property that $\text{span}(v_1, \dots v_k) = \text{im}(Q)$. So $X$ is a partial isometry from $\text{im}(Q)$ to $\text{im}(P)$ such that $X^{\ast}$ is a partial isometry the other way around, and that's why composing works. This also explains the identities above.