Simplicial approximation iff $x \in |K|$ and $f(x) \in s^{\circ} \Rightarrow |\psi|(x) \in s$ From Rotman's Algebraic Topology:

Prove that a simplicial map $\psi : K \rightarrow L $ is a simplicial approximation to $f : |K| \rightarrow |L|$ if and only if, whenever $x \in |K|$ and $f(x) \in s^{\circ}$, (where $s$ is a simplex of $L$), then $|\psi|(x) \in s$.

I'm having difficulty figuring out how to show either direction of this.  
For sufficiency, I can see that $y \in f(\text{st}(p)) \implies y = f(x)$ for some $x \in \text{st}(p) \implies y \in s^{\circ} \implies |\psi|(x) \in s$.  But I'm having trouble seeing how to show that $\psi(p) \in \text{Vert}(s).$
For necessity, I'm not seeing the connection between $f(\text{st}(p)) \subset \text{st}(\psi (p))$ and $|\psi|(x)$.
Anyone have any suggestions?
 A: 
Definition: Let $K$, $L$ be simplicial complexes and $f: \lvert K \rvert \to \lvert L \rvert$ a continuous map. A simplicial map $\psi : K \to L$ is a  simplicial approximation to $f$ if
$$ f(\text{St}_K(v)) \subset \text{St}_L(\psi(v)) $$
for any vertex $v \in K$.

Let first $\psi$ be a simplicial approximation to $f$. Let $x \in \lvert K \rvert$. Then $x$ is in the interior of a unique simplex $\sigma \in K$. Let $v \in \sigma$ be a vertex of that simplex. Then $x \in \text{St}_K(v)$ and by simplicial approximation condition $f(x) \in \text{St}_L(\psi(v))$. But if $\tau \in L$ is the unique simplex such that $f(x)$ lies in its interior, then it follows that $\psi(v)$ is a vertex of $\tau$.
This shows that $\psi$ maps the vertices of $\sigma$ to the vertices of $\tau$ and so in particular for $x \in \sigma$ we have $\lvert \psi \rvert (x) \in \tau$.

Now we prove the converse. Let $v \in K$ be a vertex and let $x \in \text{St}_K(v)$. Let $\sigma \in K$ be the unique simplex with $x$ in the interior and let $\tau \in L$ be the unique simplex with $f(x)$ in the interior. By the problem condition, we have that $\lvert \psi \rvert (x) \in \tau$. But $\lvert \psi \rvert$ is a realisation of a simplicial map $\psi$ $-$ if the vertices of $\sigma$ are $v_1, \ldots, v_k$, and $x = \sum_{i=1}^kt_iv_i$ (where all $t_i$ are positive, because $x$ is in the interior), then $$\lvert \psi \rvert (x)=\sum_{i=1}^kt_i\psi(v_i) \in \tau$$
This means that $\psi(v) \in \tau$ is a vertex of $\tau$ in particular and so $f(x) \in \text{St}_L(\psi(v))$ as required.

Remark: I like to think of stars as open neighbourhoods and my intuition is that if something is in the star, it is close to something else.
