Generating a subspace sigma-algebra Say X is a set and $\mathcal{E}$ is a collection of its subsets. 
Let $\mathcal{M}$ be the sigma-algebra generated by $\mathcal{E}$ (or the smallest sigma-algebra over X containing all members of $\mathcal{E}$)
Provide the subset $Y \subset X$ the sigma-algebra $\mathcal{M}^* = \{ M \cap Y : M \in \mathcal{M}\}$. Analogously define the collection $\mathcal{E}^* = \{E \cap Y : E \in \mathcal{E}\}$ of subsets of Y.
Is it true that $\mathcal{E}$* generates $\mathcal{M}$* (that $\mathcal{M}$* is the smallest sigma-algebra over Y containing $\mathcal{E}$*)?
EDIT:
So I found a solution. The answer is yes. It involved taking  the sigma-algebra $\mathcal{N}^*$ over Y actually generated by $\mathcal{E}^*$ and constructing the sigma-algebra $\mathcal{N}$ over X consisting of all subsets of X whose intersection with Y was a member of $\mathcal{N}^*$, noticing that $\mathcal{N}$ contained $\mathcal{E}$ and therefore contains $\mathcal{M}$, and deducing that $\mathcal{N}^* \supset \mathcal{M}^*$. The reverse inclusion is easy.
 A: The $\sigma$-algebra $\mathcal{M}^{\ast}$ can be viewed as the initial
$\sigma$-algebra generated by the inclusion map.
More precisely, let $i:Y\rightarrow X$ be the inclusion map defined
by $i(x)=x$. We can verify that $\mathcal{M}^{\ast}$ is the smallest
$\sigma$-algebra on $Y$ such that $i$ is $\mathcal{M^{\ast}}/\mathcal{M}$-measurable
and $\mathcal{M^{\ast}}$is explicitly given by $\mathcal{M}^{\ast}=i^{-1}(\mathcal{M}):=\{i^{-1}(B)\mid B\in\mathcal{M}\}$.
Also note that $\mathcal{E}^{\ast}=i^{-1}(\mathcal{E})$. In the following,
we show that $\sigma\left(i^{-1}(\mathcal{E})\right)=i^{-1}\left(\sigma(\mathcal{E})\right)$.
Since $\mathcal{E}\subseteq\sigma(\mathcal{E})$, we have $i^{-1}(\mathcal{E})\subseteq i^{-1}(\sigma(\mathcal{E}))$.
Observe that $i^{-1}(\sigma(\mathcal{E}))$ is a $\sigma$-algebra,
so $\sigma\left(i^{-1}(\mathcal{E})\right)\subseteq i^{-1}(\sigma(\mathcal{E}))$.
To show the reverse inclusion $i^{-1}(\sigma(\mathcal{E}))\subseteq\sigma\left(i^{-1}(\mathcal{E})\right)$,
we let $\mathcal{N}=\{B\subseteq X\mid i^{-1}(B)\in\sigma\left(i^{-1}(\mathcal{E})\right)\}$.
By direct verification, $\mathcal{N}$ is a $\sigma$-algebra on $X$
and $\mathcal{E}\subseteq\mathcal{N}$. Therefore $\sigma(\mathcal{E})\subseteq\mathcal{N}$
and hence $i^{-1}(\sigma(\mathcal{E}))\subseteq\sigma\left(i^{-1}(\mathcal{E})\right)$.
We remark that the theorem $\sigma\left(i^{-1}(\mathcal{E})\right)=i^{-1}\left(\sigma(\mathcal{E})\right)$
continues to hold if $Y$ is replaced by an arbitrary set and $i$
is replaced by an arbitrary map $f:Y\rightarrow X$.
