# Thread: extension of a surjective map

1. ## extension of a surjective map

Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

At first i thought this was false. But then i thought it was true... and now i've really confused myself. Can anyone help?

2. Originally Posted by junebug5389
Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

At first i thought this was false. But then i thought it was true... and now i've really confused myself. Can anyone help?
Let $\displaystyle \hat f$ be an extension of $\displaystyle f$, i.e. $\displaystyle f\subseteq \hat f$ and $\displaystyle \hat f |_S = f$.
We are told $\displaystyle f: S\to B$ is surjective.
And we are given $\displaystyle \hat f : A\to B$.
Let $\displaystyle b\in B$ then there is $\displaystyle s\in S$ so that $\displaystyle f(s) = b$.
But since $\displaystyle s\in S\subseteq A$ it means $\displaystyle \hat f (s) = \hat f|_S (s) = f(s) = b$.
Thus, $\displaystyle \hat f$ is surjective.

3. Originally Posted by junebug5389
Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

Here is another way to see it. Every function is a set of ordered pairs.
If $\displaystyle f$ is onto then $\displaystyle B = \left\{ {y:\left( {\exists x} \right)\left[ {\left( {x,y} \right) \in f} \right]} \right\}$.
So clearly any extension of $\displaystyle f$ looses no pairs.