Is $T = \{(f, f (0)) : f \in \mathbb{R^{\mathbb{R}}}\}$ a surjective function?

Question

let $T$ be a relation from $\mathbb{R^{\mathbb{R}}}$ to $\mathbb{R}$

$$T = \{(f, f (0)) : f \in\mathbb{R^{R}}\}$$

prove/ disprove that $T$ is a function. if it is - is $T$ surjective? is $T$ one to one?

I think T is a function, because for every $f$, $(f, y) \in T$ and $(f, y_1) \in T$, then $y = y_1 = f(0)$ (otherwise it's not $\in T$), but what about the surjectivity and one to one? I'm not sure how to answer that.

Answer 1

$T$ is absolutely a function, since if $f = g$ on all their inputs, the also clearly $f(0) = g(0)$.

$T$ is one to one if equal outputs imply equal inputs. And $f(0) = g(0)$ doesn't imply that the functions are equal on some non-zero value.

$T$ is onto (surjective) if the output space is covered. And there are plenty of functions with the property that $f(0) = x_0$ for any $x_0 \in \mathbb{R}$

Answer 2

Yes, $T$ is a function, for precisely the reason you gave.

To say $T$ is one-to-one is to say that, given $(f, y), (g, y) \in T$, then $f = g$. Note that $(f, y), (g, y) \in T$ if and only if $f(0) = g(0) = y$. Does $f(0) = g(0) = y$ imply that $f = g$?

To say $T$ is surjective is to say, given any particular $y_0 \in \mathbb{R}$, there is some function $f \in \mathbb{R}^\mathbb{R}$ such that $f(0) = y_0$. Can you think of such a function? (Any such function will do!)

Answer 3

so T is a function because for every f, (f,y) ∈ T and (f,y1) ∈ T, then y = y1 = f(0). T is also surjective because there many functions that exist: $f(0)$ = $x_0$ for any $x_0$ $\in$ $\mathbb{R}$.

T isn't one to one because $f (0)$ = $g(0)$ doesn't imply that for every x, $f(x) = g(x)$.