Gelfand transform on functions The Gelfand transformation identifies function spaces $C_0(X)$ for locally compact Haussdorff $X$ with commutative $C^*$ Algebras. Additionally there is a statement that if $f: X \to Y$ is a proper and continuous map, this induces a $*$-morphism $f_*: C_0(Y) \to C_0(X)$ via $f_*(g) = g \circ f$.
The condition that the map be proper is needed for the induced map to be well defined, for example if $X$ is not compact then a constant map $f: X \to Y$ will not send $C_0$ functions to $C_0$ functions. On the other hand the proof that $f_*(g)$ is in $C_0(X)$ if $g \in C_0(Y)$ is rather easy: if $\epsilon >0$, $K$ compact so that $g(Y-K)\subset B_\epsilon(0)$ then $f^{-1}(K)$ is compact and
$$f_*(g)(X-f^{-1}(K))=(g\circ f)(X-f^{-1}(K))\subset g(Y-K) \subset B_\epsilon(0)$$
There is another direction, given two commutative $C^*$ Algebras $A, B$ so that $A \cong C_0(X)$ and $B \cong C_0(Y)$ there is a statement that a proper $*$-morphism $\varphi: A \to B$ induces a proper continuous map $\varphi_*: Y \to X$. A $*$-morphism is called proper if it maps approximate identities to approximate identities.
Can somebody tell me how this map can be constructed (and why proper-ness is needed)? Also can somebody show me an example of a non-proper $*$-morphism that isn't trivial (the zero map)?
 A: For an example of a non-proper $*$-morphism that isn't trivial, consider $A=B=\mathbb C\oplus \mathbb C$, and $\phi(a,b)=(a,0)$. 
For the existence of $\varphi_*$, the key property is that for any $*$-morphism $\varphi:C_0(X)\to C_0(Y)$, there exists $\varphi_*:Y\to X$, continuous, such that $$\tag{1}\varphi(f)(y)=f(\varphi_*(y))\ \ \ \ \text{ for all }y\in Y,\ f\in C_0(X).$$ 
Note first that we may assume that $X$ and $Y$ are compact; indeed, for locally compact $X$ and $Y$, if $\tilde X$ is the one-point compactification, there is a unique correspondence between unital $*$-morphisms $C(\tilde X)\to C(\tilde Y)$ and proper $*$-morphisms $C_0(X)\to C_0(Y)$.
Thus, assume that $X$ and $Y$ are compact, and that $\varphi$ is unital.
Given a fixed $y\in Y$, define a map $\psi_y:C(X)\to\mathbb C$ by
$$
\psi_y(f)=(\varphi f)(y).
$$
It is straightforward to check that $\psi_y$ is a $*$-homomorphism (i.e., a character). The characters are given precisely by the point evaluations (this follows directly from the fact that the kernels of the characters are the maximal ideals). Because $\varphi$ is unital, $\psi_y$ is not identically zero, and so there exists an element $x\in X$ with $\psi_y(f)=f(x)$ for all $f\in C(X)$. Such an $x$ is unique, because if $x_1$ and $x_2$ satisfy $f(x_1)=\psi_y(f)=f(x_2)$, and so $x_1=x_2$ because the continuous functions separate points. We write $\varphi_*(y)$ for this element. 
We have the relation $(1)$. Now it is easy to check that $\varphi_*$ is continuous. If $y_j\to y$, then for any $f\in C(X)$
$$
f(\varphi_*(y_j))=(\varphi f)(y_j)\to(\varphi f)(y)=f(\varphi_*(y)). 
$$
As this happens for any $f\in C(X)$, we deduce that $\varphi_*(y_j)\to \varphi_*(y)$, and $\varphi_*$ is continuous. 
It remains to show that $\varphi_*$ is proper. But because we have defined it as a map between compact sets (the one-point compactifications of $Y$ and $X$), we trivially have that $(\varphi_*y)^{-1}(K)$ is compact for every compact $K\subset X$.
