homomorphism from $C (Y)$ into $C (X)$ Let $t$ be a ( ring ) homomorphism from $C (Y)$ or $C^{*} (Y)$ into   $C (X)$ . 
Can someone help me to solve the following problems?

1 : t r = r . t 1 ( r  and 1 are constant function) for each $r \in \mathbb{R}$. 
[ for each $ x \in X$, the mapping    r $\rightarrow$ (t r)(x) is a homomorphism from  or into     , and hence is either the zero homomorphism or the identity. Also,  (t1)(x)= 0  or 1.]
2:$t$ is an algebra , i.e.,  t(rg) = r.tg  $\forall r \in \mathbb{R}$  and $ g \in C (Y)$

The set $  C= C ( X ) $ of all continuous, real valued functions on a topological space$ X$  will be provided with an algebraic structure and an order structure.
The subset $  C^{*}= C^{*} ( X ) $ of $ C ( X ) $, consisting of all bounded function in $ C ( X ) $.
 A: Note that $C(X)$ (and $C^\ast(X)$ as well) has a lattice structure. It's partially ordered by pointwise ordering and $f \lor g= \max(f,g)$ and $f \land g = \min(f,g)$ is in $C(X)$ if $f,g$ are, and likewise for bounded functions.
A ring homomorphism $C(Y) \to C(X)$ is also a lattice morphism: it respects order and $\lor$, $\land$ as well.
We need the lemma that 

a ring isomorphism $f: \mathbb{R} \to \mathbb{R}$ is either $0$ or the identity. 

Proof of the lemma: $f$ sends squares to squares so $x\ge 0$ implies $x=y^2$ for some $y$, so $f(x) = f(y)^2$ hence $f(x) \ge 0$. So $f$ preserves order:
$x \le y$ iff $y-x  \ge 0$ iff $f(y-x) \ge 0$ iff $f(y) - f(x) \ge 0$ iff $f(x) \le f(y)$. A non-zero ring morphism must have $f(1) = 1$ (let $f(x) \neq 0$, then $f(x) = f(x\cdot 1) = f(x)f(1)$ and we divide by $f(x)$), and this means that $(q) = q$ for all $q \in \mathbb{Q}$ and as $f$ must be continuous (as an order preserving function), $f$ must be the identity everywhere.
As to the first question: 
Denote by $\mathbf{r}$ the constant function with value $r\in \mathbb{R}$. These functions form a subring (isomorphic to $\mathbb{R}$ by $r \to \mathbf{r}$) of $C(Y)$ and of $C^\ast(Y)$ as well. So fix $x \in X$. Then $h_x: r \to t(\mathbf{r})(x)$ is a ring morphism from the reals to itself (as every point evaluation is a ring morphism). So for every $x \in \mathbb{R}$ either $\forall r: t(\mathbf{r})(x) = 0$ or $\forall r: t(\mathbf{r})(x) = r$.
Also $t(\mathbf{1})$ has the property that $t(\mathbf{1})^2=t(\mathbf{1})$, so for all $x$: $t(\mathbf{1})(x) \in \{0,1\}$.
So essentially the function $t(\mathbf{1})$ determines the value of $t(\mathbb{r})$ as well: if $t(\mathbf{1})(x) = 0$ we have that for that $x$, for all $r$: $t(\mathbf{r})(x) = 0$ (as $r=1$ rules out the other option) and so $t(\mathbf{r})(x) = (r\cdot t(\mathbf{1}))(x)$, while $t(\mathbf{1})(x) = 1$ implies that for all $r$ we have $t(\mathbf{r})(x) = r$ and so $t(\mathbf{r})(x) = (r\cdot t(\mathbf{1}))(x)$ as well, and as this holds for all $x$, we have equality of functions and $t(\mathbf{r}) = r\cdot t(\mathbf{1})$
FInally as a corollary:
$$t(\mathbf{r}g) = t(\mathbf{r})t(g) = (r\cdot t(\mathbf{1}))t(g) = r\cdot t(\mathbf{1} \cdot g) = r\cdot t(g) = \mathbf{r} \cdot t(g)\text{.}$$
