Bundle isomorphisms for $J$-holomorphic tangent bundle In Chapter 14 of Lectures on Symplectic Geometry by da Silva, she claims that, if $(M,J)$ an almost complex manifold, then there are real bundle isomorphisms
\begin{align*}\pi_{1,0}:TM\otimes\mathbb C&\to T_{1,0}=i\text{-eigenspace of}~J\\v&\mapsto\frac12(v\otimes1-Jv\otimes i)\end{align*}
and
\begin{align*}\pi_{0,1}:TM\otimes\mathbb C&\to T_{0,1}=-i\text{-eigenspace of}~J\\v&\mapsto\frac12(v\otimes1+Jv\otimes i).\end{align*}
Does she mean for the domains to be $TM$ instead?
Also, she goes on to say that $\pi_{1,0}\circ J=i\pi_{1,0}$ and $\pi_{0,1}\circ J=-i\pi_{0,1}$ imply that there are isomorphisms of complex vector bundles
$$(TM,J)\simeq T_{1,0}\simeq\overline{T_{0,1}}.$$ I don't really understand why that's true. Maybe the most basic thing is, what does $(TM,J)$ mean as a complex vector bundle?
 A: Yes, the domains of $\pi_{1,0}$ and $\pi_{0,1}$ should be $TM$, not $TM\otimes\mathbb{C}$. This is the case for the version on her website; see here, page 78.
Note that $TM$ is a real vector bundle, and it becomes a complex vector bundle by defining $(a + bi)\cdot v := av + bJ(v)$. This is what we mean when we say $(TM, J)$ is a complex vector bundle.
With this in mind, we have $\pi_{1,0} : TM \to T_{1,0}$ and
$$\pi_{1,0}(i\cdot v) = \pi_{1,0}(J(v)) = (\pi_{1,0}\circ J)(v) = i\pi_{1,0}(v) = i\cdot\pi_{1,0}(v).$$
So the real vector bundle isomorphism $\pi_{1,0}$ is also complex linear, and hence is a complex vector bundle isomorphism.
On the other hand, we have $\pi_{0,1} : TM \to T_{0,1}$ and
$$\pi_{0,1}(i\cdot v) = \pi_{0,1}(J(v)) = (\pi_{0,1}\circ J)(v) = -i\pi_{0,1}(v) = i\cdot\pi_{0,1}(v)$$
where the action in the last expression is using the action of $\mathbb{C}$ on $\overline{T_{0,1}}$ (where $i$ acts as multiplication by $-i$). So $\pi_{0,1} : TM \to \overline{T_{0,1}}$ is an isomorphism of complex vector bundles.
Therefore $TM \cong T_{1,0} \cong \overline{T_{0,1}}$.
