Vakil 3.2.P - Morphism between Spec induced by Ring Morphism I'm currently working my way through Vakil's exercise 3.2.P, which goes as follows

Suppose $k$ is a field, $f_1,...,f_n\in k[x_1,...,x_m]$ are given. Let $\phi: k[y_1,...,y_n]\mapsto  k[x_1,...,x_m]$ be the ring morphism given by $y_i\mapsto f_i$.
(a) Show that $\phi$ induces a map of sets $\operatorname{Spec}k[x_1,...,x_m]/I\rightarrow\operatorname{Spec}k[y_1,...,y_n]/J$ for any ideals $I\subset k[x_1,...,x_m]$ and $J\subset  k[y_1,...,y_n]$
(b) Show that the map of part (a) sends the points $(a_1,...,a_m)\in k^m$ (or more precisely, $[(x_1-a_1,...,x_m-a_m)]\in\operatorname{Spec} k[x_1,...,x_m]$) to
$(f_1(a_1,..,a_m),...,f_n(a_1,...,a_m))\in k^n$.


I think I know how to prove part (a). This comes from applying this fact (which is fairly easy to prove):

Fact 1: If $\phi:B\rightarrow A$ is a map of rings, and $\mathfrak{p}$ is a prime ideal of $A$, show that $\phi^{-1}(\mathfrak{p})$ is a prime ideal of $B$.

What I'm struggling with is part (b). So, suppose I have some prime ideal $(a_1,...,a_m)$ (or, more accurately,  $[(x_1-a_1,...,x_m-a_m)]\in\operatorname{Spec}k[x_1,...,x_m]$).
Obviously by the induced map in part (a), it will be sent to the prime ideal $\phi^{-1}((a_1,...,a_m))\in k^{n}$. Except I have trouble seeing how  $\phi^{-1}((a_1,...,a_m))=(f_1(a_1,...,a_m),...,f_n(a_1,...,a_m))$.
I guess I also don't understand how $\phi$ acts on the above... for instance, is it the case that $\phi:f_1(a_1,...,a_m)\mapsto f_1$? I'm not really sure what this means to be honest.
There's probably something obvious that I'm missing at the moment... can somebody help (or have I misunderstood something crucial?)
 A: First, I'll try to comment on something you've written above, which may clarify some important details on how to approach the problem. You write that you don't "understand how $\varphi$ acts", and then (if I understand you) ask if $\varphi$ maps $f_{i}(a_{1}, \ldots, a_{m})$ to $f_{i}$. This is not the case; $f_{i}(a_{1}, \ldots, a_{m})$ is an element of $k$, and $\varphi$ preserves elements of $k$. Let me try to clarify what the definition of $\varphi$ is. 
The point is that for any $k$-algebra $A$, to specify a $k$-algebra homomorphism $k[y_{1}, \ldots, y_{n}] \to A$ is to choose $n$ elements of $A$, namely the images of $y_{1}, \ldots, y_{n}$. Explicitly, given $a_{1}, \ldots, a_{n} \in A$, there is a unique $k$-algebra homomorphism $k[y_{1}, \ldots, y_{n}] \to A$ sending $y_{i} \to a_{i}$. Indeed, one can check that the map which sends $f(y_{1}, \ldots, y_{n}) \to f(a_{1}, \ldots, a_{n})$ is a morphism of $k$-algebras (i.e., a ring homomorphism which fixes $k$) which sends $y_{i}$ to $a_{i}$, so there is at least one such morphism. On the other hand, any such morphism must send the monomial $\alpha y_{1}^{r_{1}}\cdots y_{n}^{r_{n}}$ to $\alpha a_{1}^{r_{1}} \cdots a_{n}^{r_{n}}$ (where $\alpha \in k$) by multiplicativity, and so additivity guarantees $f(y_{1}, \ldots, y_{n})$ must map to $f(a_{1}, \ldots, a_{n})$. Hence, there is exactly one such morphism for each $n$-tuple of elements of $A$; put another way, as sets, $\mathrm{Hom}_{k\mathrm{-alg}}(k[y_{1}, \ldots, y_{n}], A) \cong A^{n}$. 
Hence, Vakil defines the morphism $\varphi \colon k[y_{1}, \ldots, y_{n}] \to k[x_{1}, \ldots, x_{m}]$ by choosing elements $f_{1}, \ldots, f_{n} \in k[x_{1}, \ldots, x_{m}]$ to be the images of $y_{1}, \ldots, y_{n}$ respectively. To be explicit, as noted above, $\varphi$ sends the monomial $\alpha y_{1}^{r_{1}}\cdots y_{n}^{r_{n}}$ to $\alpha f_{1}^{r_{1}} \cdots f_{n}^{r_{n}}$.  
As for a solution to the problem, here are some hints to proceed, broken into steps (some of which you may already know):
$(1)$ Show that for any $\alpha_{1}, \ldots, \alpha_{n} \in k$, we have 
$$\{g \in k[y_{1}, \ldots, y_{n}] \mid g(\alpha_{1}, \ldots, \alpha_{n}) = 0\} = \langle y_{1} - \alpha_{1}, \ldots, y_{n} - \alpha_{n} \rangle.$$ 
(Put into words, this says that $\langle y_{1} - \alpha_{1}, \ldots, y_{n} - \alpha_{n} \rangle$ is the kernel of the evaluation morphism $k[y_{1}, \ldots, y_{n}] \to k$ which sends $g$ to $g(\alpha_{1}, \ldots, \alpha_{n})$.)
$(2)$ Show that for any $\alpha_{1}, \ldots, \alpha_{n} \in k$, $\langle y_{1} - \alpha_{1}, \ldots, y_{n} - \alpha_{n} \rangle$ is a maximal ideal of $k[y_{1}, \ldots, y_{n}]$. (Use $(1)$.)
$(3)$ Observe that $\varphi(y_{i}-f_{i}(a_{1}, \ldots, a_{m})) = f_{i} - f_{i}(a_{1}, \ldots, a_{m})$ belongs to the ideal $\langle x_{1}-a_{1}, \ldots, x_{m}-a_{m} \rangle$. (Use $(1)$.) 
$(4)$ Conclude that $\varphi^{-1}(\langle x_{1}-a_, \ldots, x_{m}-a_{m} \rangle)$ contains $\langle y_{1} - f_{1}(a_{1}, \ldots, a_{m}), \ldots, y_{n} - f_{n}(a_{1}, \ldots, a_{m})\rangle$, and so they must be equal. (Use $(2)$.) 
