Ch.7 Isomorphisms

The function $f:\mathbb{Z}\rightarrow\{0,1,2\}$ defined by $f(x)$ is the remainder when $x$ is divided by $3$ is surjective

The function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(x)=2x$ is bijective

The function $g:\mathbb{R}\rightarrow\mathbb{R}$ defined by $g(x)=\sqrt{x}$ is a right inverse of $f(x)=x^2$ because $(f\circ g)(x)=(\sqrt{x})^2=x$ but is not a left inverse because $(g\circ f)(x)=\sqrt{x^2}=x$ only when $x\ge 0$, not all $x\in\mathbb{R}$

Prove $\mathcal{P}_2\cong \mathbb{R}^3$ under the map
$f(a_0+a_1x+a_2x^2)=\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}$

Condition 1:
$f(a_0+a_1x+a_2x^2)=f(b_0+b_1x+b_2x^2)$
$\implies\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}=\begin{pmatrix}b_0\\b_1\\b_2\end{pmatrix}$
$\implies a_0=b_0,\space a_1=b_1,\space a_2=b_2$
$\rightarrow a_0+a_1x+a_2x^2=b_0+b_1x+b_2x^2$
Condition 2:
For any $\vec{w}=\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}\in\mathbb{R}^3$, the corresponding element of the domain is $a_0+a_1x+a_2x^2$ which is part of the domain
Condition 3:
$f((a_0+a_1x+a_2x^2)+(b_0+b_1x+b_2x^2))\\ =f((a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2)\\ =\begin{pmatrix}a_0+b_0\\a_1+b_1\\a_2+b_2\end{pmatrix}\\ =\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}+\begin{pmatrix}b_0\\b_1\\b_2\end{pmatrix}\\ =f(a_0+a_1x+a_2x^2)+f(b_0+b_1x+b_2x^2)$
Condition 4:
$f(r\cdot(a_0+a_1x+a_2x^2))=f(ra_0+ra_1x+ra_2x^2)\\ =\begin{pmatrix}ra_0\\ra_1\\ra_2\end{pmatrix}\\ =r\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}\\ =rf(a_0+a_1x+a_2x^2)$
Thus all four conditions are proved and $f$ is an isomorphism. Therfore, $\mathcal{R}_2\cong\mathbb{R}^3$

The function $f:\mathbb{R}^2\to\mathbb{R}^2$
$f(\begin{pmatrix}x\\y\end{pmatrix})=\begin{pmatrix}x^2\\y^2\end{pmatrix}$
does not preserve addition since
$f(\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix})=\begin{pmatrix}9\\0\end{pmatrix}\ne f(\begin{pmatrix}1\\0\end{pmatrix})+f(\begin{pmatrix}2\\0\end{pmatrix})$

$d_s:\mathbb{R}^2\to\mathbb{R}^2, \text{ multiply by nonzero scalar } s$
Maps space of $\mathbb{R}^2$ to itself, but is a dilation

$t_\theta:\mathbb{R}^2\to\mathbb{R}^2, \text{ rotate by }\theta$
Maps space of $\mathbb{R}^2$ to itself, but rotates the vector

$f_l:\mathbb{R}^2\to\mathbb{R}^2, \text{ reflect over line } l$
Maps space of $\mathbb{R}^2$ to itself, but flips the vector over a line

Consider an isomorphism $f:V\to W$ with $\vec{v}\in V$
$f(\vec{0}_V)=f(0\cdot\vec{v})=0\cdot f(\vec{v})=\vec{0}_w$

Reflexivity: space is isomorphic to itself by the identity map
Symmetry: the inverse of an isomorphism is also an isomorphism
Transitivity: evaluate the isomorphism conditions

First prove isomorphic $\implies$ same dimension
We prove that if $f:V\to W$ is an isomorphism and the basis for $V$ is $B=\langle\vec{\beta}_1,...,\vec{\beta}_n\rangle$ then its image $D=\langle f(\vec{\beta}_1),...,f(\vec{\beta}_n)\rangle$ is the basis of the codomain $W$. The reverse is equivalent by simply taking $f^{-1}$ instead.
Fix any $\vec{w}\in W$. Since $f$ is onto there must exist a $\vec{v}\in V$ s.t. $f(\vec{v})=\vec{w}$. Express $f(\vec{v})$ as $f(v_1\vec{\beta}_1+\cdots+v_n\vec{\beta}_n)=v_1f(\vec{\beta}_1)+\cdots+v_nf(\vec{\beta}_n)$ so $D$ spans $W$.
For linear independence,
$\vec{0}_W=c_1f(\vec{\beta}_1)+\cdots+c_nf(\vec{\beta}_n)=f(c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n)$
Since $f$ is one-to-one, only $f(\vec{0}_V)=\vec{0}_W$, and since $B$ is linearly independent, all $c_i$'s must be 0.

Next, prove same dimension $\implies$ isomorphic.
We prove that all $n$-dimensional spaces are isomorphic to $\mathbb{R}^n$. Then by transitivity, the statement will be proven.
The map
$\vec{v}=v_1\vec{\beta}_1+\cdots+v_n\vec{\beta}_n\xmapsto{\text{Rep}_B}\begin{pmatrix}v_1\\\vdots\\v_n\end{pmatrix}$
This function is one-to-one because if
$\text{Rep}_B(u_1\vec{\beta}_1+\cdots+u_n\vec{\beta}_n)=\text{Rep}_B(v_1\vec{\beta}_1+\cdots+v_n\vec{\beta}_n)$
then
$\begin{pmatrix}u_1\\\vdots\\u_n\end{pmatrix}=\begin{pmatrix}v_1\\\vdots\\v_n\end{pmatrix}$
which implies $u_i=v_i$ for $i=1,...,n$, implying $u_1\vec{\beta}_1+\cdots+u_n\vec{\beta}_n=v_1\vec{\beta}_1+\cdots+v_n\vec{\beta}$
The function is onto because every
$\vec{w}=\begin{pmatrix}w_1\\\vdots\\w_n\end{pmatrix}$
is an image of $\vec{v}\in V$, namely $\vec{w}=\text{Rep}_B(w_1\vec{\beta}_1+\cdots+w_n\vec{\beta}_n)$
The function preserves structure because
$\begin{array}{rcl} \text{Rep}_B(r\cdot\vec{u}+s\cdot\vec{v})& =&\text{Rep}_B((ru_1+sv_1)\vec{\beta}_1+\cdots+(ru_n+sv_n)\vec{\beta}_n)\\ &=&\begin{pmatrix}ru_1+sv_1\\\vdots\\ru_n+sv_n\end{pmatrix}\\ &=&r\begin{pmatrix}u_1\\\vdots\\u_n\end{pmatrix}+s\begin{pmatrix}v_1\\\vdots\\v_n\end{pmatrix}\\ &=&r\cdot\text{Rep}_B(\vec{u})+s\cdot\text{Rep}_B(\vec{v}) \end{array}$
So $\text{Rep}_B$ is an isomorphism, so any $n$-dimensional space is isomorphic to $\mathbb{R}^n$