An affine map is an affine isomorphism iff its underlying linear map is an isomorphism

up:: Affine Isomorphism

Transclude of Affine-Map.excalidraw

Let $(E,V,{+}_E)$ and $(E^{\prime },V^{\prime },{+}_{E^{\prime }})$ be Affine Spaces¹. Let $f:E\to E^{\prime }$ be an Affine Isomorphism with underlying linear map $L:V\to V^{\prime }$.

Affine map is bijective $⟹$ Underlying map is isomorphism

Let $f:E\to E^{\prime }$ be an affine isomorphism (Affine Map which is a Bijection). Then, for any $A\in E,v\in V$, we have

$$f(A+v)=f(A)+L(v)$$

Since $f$ has an inverse, we have that
$A+v=f^{-1}(f(A)+L(v))$
Since the inverse is also an affine map, we have that

$$A+v=A+\tilde{L}(L(v))$$

where $\tilde{L}:V^{\prime }\to V$ is the underlying linear map associated to $f^{-1}$.

However, note that, since an affine space is determined by a Regular Group Action, then there is one and only one vector connecting $A$ and $A+v$, which is $v$. Thus,

$$\forall v\in V,\tilde{L}(L(v))=v⟹\tilde{L}=L$$

And, thus, the underlying linear map of an affine isomorphism is a Vector Space Isomorphism.

Affine map is not bijective $⟹$ Underlying map is not an isomorphism

Let $f:E\to E^{\prime }$ not be an affine isomorphism; then it is not Injective or it’s not Surjective (or neither).

Suppose $f$ is not injective. Then there are $A\ne B\in E\mid f(A)=f(B)$, in which case $\overrightarrow{AB}\ne 0$ but $\overrightarrow{f(A)f(B)}=0$. Thus, $L$ is not an isomorphism, since it does not take an origin to an origin (i.e. it’s not injective).

Suppose $f$ is not surjective. Then there are points in $E^{\prime }$ which are not reached by $f$.
Let $P\in E^{\prime }\setminus f(E)$. Then there are points in $E^{\prime }$ which have a distance of, say, $w\in V^{\prime }$ from $P$ and which are still in $E^{\prime }$. ?????

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References

• Notes on Mathematical Physics for Mathematicians - Daniel Tausk

Footnotes

1. $V^{\prime }$ is just a vector space (possibly) unrelated to $V$. Not to be confused with $V$‘s dual. ↩