$j''(X,\mathcal{B})$ being a subspace of $(j(X),j(\mathcal{B}))$ for elementary $j$. Let $X$ be a topological space with basis $\mathcal{B}$ such that every point has a local basis of size $<\kappa$, and let $j:V\rightarrow M$ be elementary with $crit(j)=\kappa$. Denote $j''(X,\mathcal{B})$ the topological space $(j''X,\{j''B:B\in\mathcal{B}\})$, I'm trying to verify that this is a subspace of $(j(X),j(\mathcal{B}))$. 
(We're considering the topologies with $\{j''B:B\in\mathcal{B}\}$ and $j(\mathcal{B})$ as bases respectively.)
A proof of this is provided in the paper New proofs of the consistency of the normal Moore space conjecture I. (very end of lemma 2.4) where I encounter the following confusion:
Given $j(x)\in j''X$ let $\{B_\alpha\}_{\alpha<\rho}$ with $\rho<\kappa$ be a local basis at $x$, then it is stated that $\{{j(B_\alpha)}\}_{\alpha<j(\rho)}$ is by elementarity a local basis at $j(x)$ in $j(X)$. I can't make sense of this.
Elementarity would give that $j(\{B_\alpha\}_{\alpha<\rho})$ is a local basis at $j(x)$ in $j(X)$ but $$j(\{B_\alpha\}_{\alpha<\rho})\neq\{{j(B_\alpha)}\}_{\alpha<j(\rho)}\quad\big(=\{{j(B_\alpha)}\}_{\alpha<\rho}=j''(\{B_\alpha\}_{\alpha<\rho})\big).$$
If what they meant was that, by elementarity, there exists a local basis at $j(x)$ in $j(X)$ of size $j(\rho)$, then I see no reason why the elements of this basis would be of the form $j(B_\alpha)$ with $B_\alpha\in \mathcal{B}$. Which is what we want as we will later use
$$j(B_\alpha)\cap j''X=j''B_\alpha$$
to conclude that $$\{j''B_\alpha\}_{\alpha<\rho}=\{j(B_\alpha)\cap j''X\}_{\alpha<j(\rho)}$$
which would mean we found a single set which is a local basis in both topologies, and so that both topologies are the same. That is, $(j''X,\{j''B:B\in\mathcal{B}\})$ is a subspace of $(j(X),j(\mathcal{B}))$.
On a side note, I don't understand the use of $j(\rho)=\rho$ other than making sense of the set $\{{j(B_\alpha)}\}_{\alpha<j(\rho)}$, which is used before any comment of $\rho=j(\rho)$ is made.
I include this possibly relevant screenshot (although there are a couple of typos and some notation I didn't define):

Thank you very much for reading.
 A: 
but $j(\{B_\alpha\}_{\alpha<\rho})\neq\{{j(B_\alpha)}\}_{\alpha<j(\rho)}$

Actually, the two are equal in this context - remember that $\rho<\kappa=crit(j)$, so $j(\rho)=\rho$. For all sequences of length below the critical point, $j$ may affect each individual term but it won't "stretch the sequence."

Let me sketch a proof of this fact, which I'll phrase precisely as: 

$(*)$ Suppose $j:V\rightarrow M$ is a nontrivial elementary embedding with $crit(j)=\kappa$. Then for each $\rho<\kappa$ and each sequence $X=(x_\eta)_{\eta<\rho}$ of length $\rho$, we have $j(X)=(j(x_\eta))_{\eta<\rho}.$

Proof. First, we note that by elementarity $j(X)$ is (in $M$) a sequence of length $j(\rho)$; since $\rho=j(\rho)$ (since $\rho<\kappa$), this tells us that $M$ thinks that $j(X)$ is a sequence of length $\rho$. "Is a sequence of length $\rho$" is absolute between $V$ and $M$, so $j(X)$ really is a sequence of length $\rho$.
Now let's think about its terms. Consider the statement "$y$ is the $\gamma$th term of the sequence $X$" (for $\gamma<\rho$). This is a first-order sentence, so by elementarity it holds iff the sentence "$j(y)$ is the $j(\gamma)$th term of the sequence $j(X)$" holds. Taking $y=x_\gamma$ and noting that $j(\gamma)=\gamma$, we have that the $\gamma$th term of $j(X)$ is $j(x_\gamma)$ for each $\gamma<\rho$ (and for $\gamma\ge\rho$, there is no $\gamma$th term as pointed out above).
