Commit 14812eef by Jim Hefferon

### graphics for slid three_ii

 ... ... @@ -720,13 +720,17 @@ consider dropping the one-to-one condition. With that, there may be some output vectors $\vec{w}\in W$ for which there are many associated inputs, $\vec{v}\in V$ such that $h(\vec{v})=\vec{w}$. This is a real difference. Now, we can ask: Now we can ask: for any vector in the range $\vec{w}\in\rangespace{h}$ what are the associated domain vectors $\vec{v}\in V$? \pause \smallskip \ExecuteMetaData[../map2.tex]{InverseImage} \smallskip The structure of the inverse image sets will give us insight into the definition of homomorphism. \end{frame} ... ... @@ -787,8 +791,9 @@ This function $\map{h}{\Re^2}{\Re^2}$ is linear. \begin{equation*} \colvec{x \\ y}\mapsto\colvec{x+y \\ 2x+2y} \end{equation*} Here are elements of $h^{-1}(\colvec{1 \\ 2})$ going to $\colvec{1 \\ 2}$. (Only one inverse image element is shown as a vector, most are shown as dots.) Here are elements of $h^{-1}(\colvec{1 \\ 2})$. (Only one inverse image element is shown as a vector, most are indicated with dots.) \begin{center} \includegraphics{asy/three_ii_inv_img01.pdf} \quad\raisebox{0.25in}{$\longmapsto$}\quad ... ... @@ -821,10 +826,14 @@ vector. That is, preservation of addition is: $h({\color{red}\vec{v}_1})+h({\color{blue}\vec{v}_2}) =h({\color{magenta}\vec{v}_1+\vec{v}_2})$. \pause \end{frame} \begin{frame} So the intuition is that a linear map organizes its domain into inverse images, such that those sets reflect the structure of the range. images, \begin{center} \includegraphics{../ch3.5} % bean to bean; many to one \end{center} such that those sets reflect the structure of the range. \end{frame} ... ...