On pg. 71 of Guillemin and Pollack they prove the $\epsilon$-Neighborhood theorem.
Here $Y$ is a compact boundaryless manifold in $\mathbb{R}^M$. They say
Proof: Let $h:N(Y)\to\mathbb{R}^M$ be $h(y,v)=y+v$. Notice that $h$ is regular at every point of $Y\times\{0\}$ in $N(Y)$, for through $(y,0)$ there pass two natural complementary manifolds of $N(Y)$, $Y\times\{0\}$ and $\{y\}\times N_y(Y)$. The derivative of $h$ at $(y,0)$ maps the tangent space of $Y\times\{0\}$ at $(y,0)$ onto $T_y(Y)$, and maps the tangent space of $\{y\}\times N_y(Y)$ at $(y,0)$ onto $N_y(Y)$.
I don't understand the last sentence "The derivative of $h$ at $(y,0)$ maps the tangent space of $Y\times\{0\}$ at $(y,0)$ onto $T_y(Y)$, and maps the tangent space of $\{y\}\times N_y(Y)$ at $(y,0)$ onto $N_y(Y)$."
Can anyone explain why that is please? Thanks.
I tried to compute out $$ dh_{(y,0)}(y_1,v)=y_1+v $$ but how does that imply $dh_{(y,0)}(T_{(y,0)}(Y\times\{0\}))=T_y(Y)$?
Note that $h$ is the restriction to $N(Y)\subset\Bbb R^M\times\Bbb R^M$ of the linear map $$H\colon\Bbb R^M\times\Bbb R^M \to \Bbb R^M, \quad H(x,w)=x+w\,,$$ so $dH_{(y,0)} = H$. Thus, $dh_{(y,0)} = H\big|_{T_{(y,0)}N(Y)}$, as you suggested. When you write $(y_1,v)\in T_{(y,0)}N(Y) = T_y(Y) \times \{0\} \oplus \{0\}\times N_y(Y)$, this means $y_1\in T_y(Y)$ and $v\in N_y(Y)$. Since $dh_{(y,0)}(y_1,0) = y_1$ for every $y_1\in T_y(Y)$, this gives the surjection you desired. And similarly, because $dh_{(y,0)}(0,v)=v$ for every $v\in N_y(Y)$, you get the other surjection.