How to view an inclusion of $k'$-rational points Let $X$ be an algebraic $k$-scheme in the sense of these notes (http://www.jmilne.org/math/CourseNotes/iAG200.pdf), and let $k'$ be a field containing $k$.  By $X(k')$, we mean the set of morphisms of $k$-schemes $\textrm{Hom}(\textrm{Max}(k'),X)$.  We say that $X(k')$ is dense in $X$ if the only closed subscheme $Z$ of $X$ for which $Z(k') = X(k')$ is $X$ itself.
I'm a little confused on how we regard $Z(k')$ as a subset of $X(k')$ in the first place.  If $X$ is affine, say $X = \textrm{Max } A$ for some finitely generated $k$-algebra $A$, then $X(k')$ is the set of $k$-algebra homomorphisms from $A$ to $k'$.  If $Z$ is a closed subscheme, then we have $Z \cong V(\mathfrak a)$ for some ideal $\mathfrak a$ of $A$, and $Z(k')$ can be identified with $\textrm{Hom}_{\textrm{k-alg}}(A/\mathfrak a, k')$.  This gives us a natural inclusion $$\textrm{Hom}_{\textrm{k-alg}}(A/\mathfrak a, k') \rightarrow \textrm{Hom}_{\textrm{k-alg}}(A, k')$$ $$f \mapsto f \circ \pi$$ which is clearly injective ($\pi$ is the projection $A \rightarrow A/\mathfrak a$).  But if $X$ is not affine, I'm having trouble seeing that the natural map $$\textrm{Hom}_{}(\textrm{Max } k', Z) \rightarrow \textrm{Hom}_{}(\textrm{Max } k', X) $$ $$ f \mapsto i \circ f$$ is injective. Here $i: Z \rightarrow X$ is a morphism of schemes which on the level of topological spaces is the inclusion map of a closed set $Z$.  Certainly if $f, g$ are morphisms of $\textrm{Max } k'$ to $Z$, then on the level of topological spaces, they are determined by where they map the single maximal ideal of $k'$.  If they don't map to the same point, then $i \circ f \neq i \circ g$.  But if $f$ and $g$ map $(0)$ to the same point, do they have to be the same morphism of $k$-schemes?
 A: I figured it out.  Let me prove a more general result in a very long winded way:
Theorem: Let $Z, X$ be algebraic $k$-schemes.  An immersion of $Z$ into $X$ is an open subscheme of a closed subscheme of $X$ (or the other way around, same thing).  If $\iota: Z \rightarrow X$ is an immersion, then $\iota_{\ast}:Z(k') \rightarrow X(k'), \phi \mapsto \iota \circ \phi$ is injective.
Lemma 1: The theorem holds when $Z$ is an open immersion of $X$.  
Obvious, because giving a morphism of schemes $\phi \in X(k')$ is the same as giving a point $x \in X$ and a monomorphism of $k$-algebras $\mathcal O_{X,x}/\mathfrak m_x \rightarrow k'$.
Lemma 2: The theorem holds when $\iota$ is a closed immersion and $X$ is affine.
In this case, we can write $X = \textrm{Max } A$ for some finitely generated $k$-algebra $A$, and $Z$ is the maximal spectrum of $A/\mathfrak a$ for some ideal $\mathfrak a$ of $A$, the immersion $Z \rightarrow X$ corresponding to a projection $\pi: A \rightarrow A/\mathfrak a$.  Here the mapping $Z(k') \rightarrow X(k')$ corresponds to the mapping $\textrm{Hom}_k(A/\mathfrak a, k') \rightarrow \textrm{Hom}_k(A,k'), f \mapsto f \circ \pi$.  This is actually a homomorphism of $k$-algebras, so we just have to show the kernel is trivial, which is obvious.
Lemma 3: Let $U_i$ be an open cover of $X$, and let $\eta_i: U_i \rightarrow X$ be the inclusion morphisms.  Then $X(k')$ is the union of the images $(\eta_i)_{\ast} [U_i(k')]$, where $\phi$ lies in the image of $(\eta_i)_{\ast}$ if $\psi$ maps the unique maximal of $k'$ to a point $U_i$.
To give a morphism $\phi$ from $\textrm{Max } k'$ into $X$ is to give a morphism into one of the $U_i$s, followed by $\eta_i$.  So this is clear.
Proof of the theorem: by the first lemma, we can assume that $Z$ is a closed immersion, and without loss of generality we can assume $Z \subseteq X$.  Given the inclusion morphism $\iota: Z \rightarrow X$, let $\iota_{\ast}$ be the corresponding map $Z(k') \rightarrow X(k')$.  Let $U_i$ be an open cover of $X$ by open affines, and let $\eta_i: U_i \rightarrow X$ be the inclusion morphism.  Each intersection $U_i \cap Z$ admits a closed immersion $\delta_i$ into $U_i$, and an open immersion $\gamma_i$ into $Z$.  Then $\iota \circ \gamma_i = \eta_i \circ \delta_i$, hence $\iota_{\ast} \circ (\gamma_i)_{\ast} = (\eta_i)_{\ast} \circ (\delta_i)_{\ast}$.  The maps $(\gamma_i)_{\ast}$ and $(\eta_i)_{\ast}$ are injective by the first lemma, and each $(\delta_i)_{\ast}$ is injective by the second lemma.  
Now our whole goal is to show that $\iota_{\ast}$ is injective.  Let $\phi, \phi' \in Z(k')$ be unequal.  We want to show that $\iota_{\ast}(\phi) \neq \iota_{\ast}(\phi')$.  If $\phi$ and $\phi'$ are distinct maps on the level of topological spaces, then this is obviously the case (as I mentioned in my original question).  Otherwise, $\phi$ and $\phi'$ map the unique maximal ideal of $k'$ to the same point, and in particular, by Lemma 3 (applied to the open covering of $Z$ by the $Z \cap U_i$), they both lie in the image of $(\gamma_i)_{\ast}$ for some $i$.  So we can write $\phi = (\gamma_i)_{\ast}(\psi), \phi' = (\gamma_i)_{\ast}(\psi')$.  Since $(\gamma_i)_{\ast}$ is injective, $\psi \neq \psi'$.  Hence $$\iota_{\ast}(\phi) = \iota_{\ast} \circ (\gamma_i)_{\ast}(\psi) = (\eta_i)_{\ast} \circ (\delta_i)_{\ast}(\psi) \neq  (\eta_i)_{\ast} \circ (\delta_i)_{\ast}(\psi')  =  \iota_{\ast} \circ (\gamma_i)_{\ast}(\psi') = \iota_{\ast}(\phi')$$
