About some terminology and notation from elementary topology I am studying real analysis where some topological concepts come up from time to time. I looked up a couple of topology related proofs where I am not clear on a few things.

Let $f: (X, \tau) \to (Y, \tau')$ be continuous. Let $g|_A$ be a restriction of $f$ to $A$ where $\emptyset \ne A \subseteq X.$ Then for any $\tau'$-open $H \subseteq Y, \ g^{-1}(H) = \{a \in A: f(a) = g(a) \in H\} = A \cap f^{-1}(H)$ which is $\tau_A$-open meaning $g$ is continuous.

Does $g$ retain the same topologies from $f$? I mean, do we have $g: (A, \tau) \to (Y, \tau')$? Also, since $f$ is continuous, $f^{-1}(H)$ is open, but how do we know $A \cap f^{-1}(H)$ is open? Almost every analysis book has a theorem about  the fact that the intersection of a finite number of open sets is open. Going by that we'd want $A$ to be open. But then the assumption $A \subseteq X$ says nothing about the openness of $A$. Either way, why is $A \cap f^{-1}(H)$ open?

Given a mapping $f: (X, \tau) \to (Y, \tau′)$ and a set $Z$ satisfying $f(X) \subseteq
Z \subseteq Y$, let $f_0$ be the same map as $f$ except that its codomain is $Z$. The $\color{red}{\text{$\tau′_Z$-open sets take the form $Z \cap H$}}$, where $H \in \tau′$. Now $f_0^{-1}(Z \cap H) = f^{-1}(H)$.

Just to be clear, are we considering $f_0: (X, \tau) \to (Z, \tau')$? Are "$\tau′_Z$-open sets" the same as open subsets of $Z$? I ask these questions just in case things work differently in topology proper :)
I think the part in red of the quote requires a proof. If $H \subset Z$, then we are done. Suppose $H \subset Y\setminus Z$. Then $Z \cap H = \emptyset$. But $\emptyset$ is an open subset of every set, so the part in red above holds. Now let $\emptyset \ne H = Z \cap Y.$ Then $Z \cap H \subseteq Z$. If $Z \cap H$ is open, we are done. How do we show it? If the foregoing is incorrect, how do they know the statement in red above holds? Also, how does the equality in the second quote above hold if $H \ne Z \cap H$? Say, when $H \ne \emptyset$ and $Z \cap H = \emptyset?$
 A: 
Does  retain the same topologies from ? I mean, do we have :(,)→(,′)?

In our case, there is no topology to assume on $A$ except for the subset topology, whose open sets are the sets $U\cap A$ for $U$s that were open in $X$.
This means that continuity $g:A\to Y$ is given when the preimages of open sets in $Y$ are sets $V\subseteq A$ such that $V=U\cap A$ for some $U\subseteq X$ open.

Either way, why is $∩^{−1}()$ open?

I guess this was also answered above?

Just to be clear, are we considering $_0:(,)\to (,′)$?

Yes

Are "$′$-open sets" the same as open subsets of ?

Yes, according to the subset topology inherited from $Y$.
As to the last question/proof, such sets are open by definition the subset topology, and it is defined this way so that any map $f: X\to Z$ is continuous iff $i\circ f:X\to Z\to Y$ is continuous, where $i:Z\to Y$ is the inclusion (identity) of $Z$ inside $Y$. (That is, $i:Z\to Y$ is given by $i(z)=z$ where $Z\subseteq Y$, and $Z$ is taken with the subset topology described above)
