extension of a surjective map

**junebug5389** · November 22nd 2008, 02:29 PM

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?

**ThePerfectHacker** · November 22nd 2008, 02:56 PM

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 $\hat f$ be an extension of $f$ , i.e. $f\subseteq \hat f$ and $\hat f |_S = f$ .
We are told $f: S\to B$ is surjective.
And we are given $\hat f : A\to B$ .
Let $b\in B$ then there is $s\in S$ so that $f(s) = b$ .
But since $s\in S\subseteq A$ it means $\hat f (s) = \hat f|_S (s) = f(s) = b$ .
Thus, $\hat f$ is surjective.

**Plato** · November 22nd 2008, 03:19 PM

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 $f$ is onto then $B = \left\{ {y:\left( {\exists x} \right)\left[ {\left( {x,y} \right) \in f} \right]} \right\}$ .
So clearly any extension of $f$ looses no pairs.

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