Invariance of subharmonicity under a holomorphic map 
If $f:U_1\rightarrow U_2$ is holomorphic and $u$ is subharmonic on $U_2$, then prove that $u\circ f $ is subharmonic on $U_1$.

I know how to prove the same argument with $f$ being conformal. In that proof I had to use $f^{-1}$. But here $f$ is just holomorphic and not necessarily invertible. So how could I prove the above statement? Any help is appreciated.
 A: I set aside the trivial case in which $u$ is identically $-\infty$.
By a well-known approximation (see, for example, by Theorem 1.IV.10 in Doob's Classical Potential Theory and it Probabilistic Counterpart) there is a decreasing sequence $(u_n)$ of functions subharmonic on $U_2$ with $\lim_n u_n(x)=u(x)$ for all $x\in U_2$, such that, in addition,  each $u_n$ is smooth and bounded below. As noted by @Hmm, a straighforward calculation of $\Delta(u_n\circ f)$ shows that $u_n\circ f$ is subharmonic on $U_1$  for each $n$. The monotone limit $u\circ f=\lim_nu_n\circ f$ is then also subharmonic; see the beginning of section 1.II.4 in Doob ibid.
A probabilistic proof can be based on the fact that if $(B_t: t\ge 0)$ is a complex Brownian motion, then $(f(B_t): 0\le t<\tau)$ is (the time change of) a Brownian motion in $U_2$; here $\tau:=\inf\{t: B_t\notin f^{-1}(U_2)\}$. From this it follows that $u\circ f(B_t)$ is a submartingale up to time $\tau$, which in turn implies the sub-mean inequality for $u\circ f$.
