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?)
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 $(2)$.)