I've been reading Harvey Rose's book on Linear Algebra, and I am stuck trying to understand this proof.
The short exact sequence is $ \{0\} \rightarrow U \rightarrow^f V \rightarrow^g W \rightarrow \{0\}$
To prove it splits, they say:
Let $\{w_1,...,w_n\}$ be a basis for W. Choose $v_i\in V$ so that $v_i$ is a preimage of $w_i$ under the map $g$, which can happen because $g$ is surjective. Now they say that there will exist a map $h: W \rightarrow V$ with the property that $h(w_i)=v_i$ for all $i$ and that $goh$ is the identity map.
Here's my problem:
I understood that there will exist a $v_i$ for each $w_i$ because $g$ is a surjection, but how does that imply that $h$ exists? Will not $g$ have to be one to one as well? Or isn't it necessary that V and W have the same dimension?
Don't forget that $\{w_1,\ldots,w_n\}$ is a basis of $W$. Therefore, for any vector space $Z$ and for any vectors $z_1,\ldots,z_n\in Z$, there is one (and only one) linear map $f\colon W\longrightarrow Z$ such that$$(\forall k\in\{1,2,\ldots,n\}):f(w_k)=z_k.$$That's all you need.