Let $X,Y$ be real vector spaces, $T: X\to Y$ be a linear map, and fix a nonempty $S\subseteq X$ (we do not assume that $S$ is neither convex nor compact (indeed, right now we do not assume any topological structure on $X$ and $Y$)). Everything below is just basic linear algebra, hence you may prefer to go directly to the final question.


A point $x \in S$ is called an *extreme point* if $x=\alpha y+(1-\alpha)z$, for some $\alpha \in (0,1)$ and $y,z \in S$, implies $y=z$. Denote the set of extreme points with $\mathrm{Ext}\_X(S)$.
The following question is related to the commutation of $T$ and the $\mathrm{Ext}\_X$ operator, cf. [this](https://math.stackexchange.com/questions/1922019/linear-map-of-extrem-points-and-its-injectivity-surjectivity?noredirect=1&lq=1) and [this one](https://math.stackexchange.com/questions/1591561/preservation-of-extreme-points-under-linear-transformation).


If $T$ is an injection then the condition $Tx=\alpha Ty+(1-\alpha)Tz$, with $x,y,z \in S$ and $\alpha \in (0,1)$, is equivalent to $x=\alpha y+(1-\alpha)z$, and similarly $Ty=Tz$ is equivalent to $y=z$. Hence $Tx$ is an extreme point of $T[S]$ if and only if $x$ an extreme point of $S$. Hence:



> 
> **Fact**. If $T \in \mathcal{L}(X,Y)$ is injective, then $\mathrm{Ext}\_Y(T[S])=T[\mathrm{Ext}\_X(S)]$.
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Now, let us suppose that $T$ is not necessarily injective.
Define $\hat{X}:=X/\mathrm{Ker}(T)$ be the quotient vector space and $\hat{T}: \hat{X}\to Y$ by $\hat{T}(x+\mathrm{Ker}(T))=Tx$ for all $x \in X$.
Then $\hat{T}$ is a well-defined linear injection.
We obtain by the Fact above (applied to $\hat{T}$ and $S+\mathrm{Ker}(T)$) that:
$$
\mathrm{Ext}\_{Y}(\hat{T}[S+\mathrm{Ker}(T)])=\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))],
$$
so that
$$
\mathrm{Ext}\_{Y}(T[S])=\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))].
$$
(As it has been observed in the linked thread, the linear map $T:\mathbf{R}^2\to \mathbf{R}$ defined by $T(x,y)=x$ is such that $\mathrm{Ext}\_Y(T[S])\subsetneq T[\mathrm{Ext}\_X(S)]$ if $S$ is the closed unit ball of $\mathbf{R}^2$ with the $1$-norm. However, the above identity is correct since $\mathrm{Ext}\_{Y}(T[S])=\mathrm{Ext}\_T([-1,1])=\{-1,1\}$ and $\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))]=\hat{T}(\mathrm{Ext}\_{\hat{X}}([-1,1]\times \mathbf{R}))=\{-1,1\}$.)



> 
> **Preliminary Question 1.** Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\mathrm{Ext}\_Y(T[S])\subseteq T[\mathrm{Ext}\_X(S)]$?
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Considering the above premises, this can be rewritten equivalently as follows:



> 
> **Preliminary Question 2.** Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))]\subseteq T[\mathrm{Ext}\_X(S)]$?
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Suppose that the left hand side is nonempty (otherwise there is nothing to prove) and pick $x+\mathrm{Ker}(T)\in \mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))$, i.e., if $\hat{T}(x+\ker{T})=\alpha \hat{T}(y+\mathrm{ker}(T))+(1-\alpha) \hat{T}(z+\mathrm{ker}(T))$, for some $\alpha \in (0,1)$, then $y-z \in \mathrm{Ker}(T)$. Equivalently, if $Tx=\alpha Ty+(1-\alpha)Tz$ then $Ty=Tz$.
We would like to show, *if* the Question has a positive answer, then $Tx \in T[\mathrm{Ext}\_X(S)]$, i.e., there exists $h \in \mathrm{Ext}\_X(S)$ such that $Tx=Th$.
However, this suggested the counterexample: pick $T:\mathbf{R}^2\to \mathbf{R}$ as the example above and $S:=\{0\}\times (0,1)$, so that
$$
\mathrm{Ext}\_Y(T[S])=\{0\}\not\subseteq \emptyset= T[\mathrm{Ext}\_X(S)]
$$
Hence, the Preliminary Questions have a negative answer. To avoid similar conterexamples with empty sets, we assume that $X,Y$ are Banach spaces, $T$ is continuous, $S$ (and so also $T(S)$) is compact convex. Hence we come to our final question:



> 
> **Question** Suppose that $X,Y$ are Banach spaces, $T$ is continuous, and $S\subseteq X$ is compact convex. Is it true that $\mathrm{Ext}\_Y(T[S])\subseteq T[\mathrm{Ext}\_X(S)]$?
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For each $x \in S$, the coset $x+\mathrm{Ker}(T)$ is closed and convex, hence $H:=(x+\mathrm{Ker}(T)) \cap S$ is nonempty compact convex. Continuing the reasoning above, the answer should be positive in Hilbert spaces, picking the unique $h=h(x) \in H$ which minimizes the distance of $H$ from the origin (and, more generally, in Banach spaces with a strictly convex norm). Is it correct also in general Banach spaces?



Commutation of linear maps and extreme points
The answer to the question is **yes**.


*Proof.* Let $y$ be an extreme point of $T[S]$. Then
$$
 F :=T^{-1}(\{y\}) \cap S
$$
is non-empty, compact and a face of $S$. By the Krein-Milman theorem, $F$ has an extreme point $x$, and since $F$ is a face of $S$, it follows that $x$ is also an etreme point of $S$. Since $Tx = y$, this proves the claim.


