Commit acc4944c authored by Jim Hefferon's avatar Jim Hefferon

final edits of map2

parent 6b7148f8
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......@@ -12012,7 +12012,7 @@
\end{equation*}
We need only to verify that this map is onto.
Any member of $W$ can be written as a linear combination of
We can write any member of $W$ as a linear combination of
basis elements
$c_1\cdot \vec{w}_1+\dots+c_k\cdot \vec{w}_k$.
This vector is the image, under the map described above, of
......@@ -12148,7 +12148,7 @@
\( r=h(c\vec{v}) \).
Thus the rank of \( h \) is one.
Because the nullity is given as $n$, the dimension of the domain of
Because the nullity is $n$, the dimension of the domain of
\( h \), the vector space \( V \), is \( n+1 \).
We can finish by showing
\( \set{\vec{v},\vec{\beta}_1,\dots,\vec{\beta}_n} \)
......@@ -12193,7 +12193,7 @@
c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n
\mapsunder{h} c_1x_1+\dots+c_nx_n
\end{equation*}
is easily seen to be linear, and to be mapped by $\Phi$ to the
is linear and $\Phi$ maps it to the
given vector in $\Re^n$, so \( \Phi \) is onto.
The map \( \Phi \) also preserves structure:~where
......@@ -12240,12 +12240,12 @@
\begin{ans}{Three.II.2.43}
Either yes (trivially) or no (nearly trivially).
If \( V \) `is homomorphic to' \( W \) is taken to mean there is a
If we take \( V \) `is homomorphic to' \( W \) to mean there is a
homomorphism from \( V \) into (but not necessarily onto) \( W \),
then every space is homomorphic to every other space as a zero map always
exists.
If \( V \) `is homomorphic to' \( W \) is taken to mean there is an
If we take \( V \) `is homomorphic to' \( W \) to mean there is an
onto homomorphism from \( V \) to \( W \) then the relation is not
an equivalence.
For instance, there is an onto
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