I'm trying to fill in the details to a line in a proof in Kanamori's exposition of Solovay's theorem on Lebesgue Measurability. I include the pertient information: 
My question is on "$\kappa$ is still inaccessible in $V[a]$ by 10.12 and 10.17(e)."
By these lemmas I know that $\kappa$ is still inaccessible in some $V[G\cap Col(\lambda,\delta)]$ that contains $V[a]$ (for this I'm assuming $|Col(\lambda,\delta)|<\kappa$ which I haven't actually checked) but I feel like I need $V[a]$ to be actually equal to some extension of this form to get $\kappa$ to be inaccessible. Sorry if this is a silly question I've been feeling very confused for a while.
Being inaccessible is downwards absolute: if $W_1$ is an end extension of $W_0$ (that is, $W_0\subseteq W_1$ and for every $a\in W_0$, $\{b\in W_1: W_1\models b\in a\}=\{b\in W_0: W_0\models b\in a\}$ - e.g. forcing extensions are end extensions, $V$ is an end extension of $L$, etc.), then every inaccessible cardinal in $W_1$ is an inaccessible cardinal in $W_0$. For example, any inaccessible cardinal in $V$ is also inaccessible in $L$.
Why? Well, first let's show that every cardinal of $W_1$ is a cardinal of $W_0$. Suppose $W_0$ thinks $\alpha$ isn't a cardinal. Then $W_0$ contains some $f$ which is a bijection between $\alpha$ and some ordinal $\beta<\alpha$. But then $f\in W_1$, since $W_0\subseteq W_1$; and so $W_1$ also sees that $\alpha$ isn't a cardinal.
The same reasoning carries us through inaccessibility. For $\lambda<\kappa$, $W_1$ thinks there is no surjection from $2^\lambda$ to $\kappa$; since $W_1\supseteq W_0$, we have $(2^\lambda)^{W_0}\subseteq (2^\lambda)^{W_1}$, and so $W_0$ thinks there is no surjection from $2^\lambda$ to $\kappa$ either. Similarly, if $f: \alpha\rightarrow\kappa$ is in $W_0$, then $f$ is in $W_1$ as well, so any witness to $\kappa$ being singular in $W_0$ lifts to a witness to $\kappa$ being singular in $W_1$. (E.g., it is impossible to force a singular cardinal to be regular, although the opposite is easy.)
Lots of large cardinal properties are downwards absolute in this way - specifically, the ones which are $\Pi_1$. However, there are also plenty which aren't; a trivial example being measurable cardinals, since there are no measurables in $L$.
Of course, the converse fails: an ordinal $\kappa\in W_1$ may seem "large" in $W_0$, while $W_1$ sees that it is actually quite small (e.g. countable). For example, if there is a measurable cardinal in $V$, the $\omega_1^L$ is countable; indeed, there are countable (in $V$) ordinals which $L$ thinks are inaccessible, Mahlo, weakly compact, . . .