How can we prove for a tychonoff space $X$ and a metric space $(Y, d)$, that a map $\phi : Y \rightarrow C(X, Y)$, is an embedding.

Question

How can we prove for a tychonoff space $X$ and a metric space $(Y, d)$, that a map $\phi : Y \rightarrow C(X, Y)$, $\phi(y) = f$, $f$ is a constant map in $C(X, Y)$ is an embedding. Where $C(X, Y)$ is equipped with uniform topology.

Answer 1

The uniform topology on $C(X,Y)$ has as a base all sets of the form $B(f,r)$ where $f \in C(X,Y)$ and $r>0$ and $$B(f,r) = \{g \in C(X,Y): \forall x \in X: d(f(x),g(x)) < r\}$$

Now $\phi$ which maps $y \in Y$ to $\phi(y) \in C(X,Y)$ defined by $\phi(y)(x)=y$ for all $x \in X$, is continuous:

Suppose $y_n \to y_0$ in $(Y,d)$ (it's enough to show sequential continuity, as $Y$ is a first countable space), then it's clear that $\phi(y_n) \to \phi(y_0)$: if $B(\phi(y_0), r)$ is a basic neighbourhood of $\phi(y_0)$, then there is some $N$ such that $n \ge N \to d(y_n, y_0) < r$ by convergence and then also $n \ge N \to \phi(y_n) \in B(\phi(y_0), r)$ and we're done.

$\phi$ is clearly 1-1 and $\phi$ is an embedding as for all $y \in Y$ and $r>0$ we have

$$\phi[ B_d(y,r)] = B(\phi(y), r) \cap \phi[Y]$$

and so $\phi$ maps open sets to relatively open subsets in $\phi[Y]$.