[proofplan] We prove both assertions directly from the two-point definition of convexity. For the image statement, two points of $T(C)$ are represented as images of two points of $C$, and affinity of $T$ moves the convex combination through $T$. For the preimage statement, two points of $T^{-1}(D)$ have images in $D$, and affinity shows that the image of their convex combination is a convex combination in $D$. [/proofplan] [step:Use affinity to move convex combinations through $T$] Since $T: V \to W$ is affine, for every $x_0, x_1 \in V$ and every $t \in [0,1]$, \begin{align*} T((1-t)x_0 + t x_1) = (1-t)T(x_0) + tT(x_1). \end{align*} This is the only property of affine maps used below. [/step] [step:Represent two points of the image by points of $C$] Assume $C \subset V$ is convex. To prove that $T(C)$ is convex, let $y_0, y_1 \in T(C)$ and let $t \in [0,1]$. By the definition of image, there exist $x_0, x_1 \in C$ such that \begin{align*} y_0 = T(x_0) \end{align*} and \begin{align*} y_1 = T(x_1). \end{align*} Since $C$ is convex and $x_0, x_1 \in C$, the point \begin{align*} x_t := (1-t)x_0 + t x_1 \end{align*} belongs to $C$. [/step] [step:Conclude that the image is convex] Using affinity of $T$ at the points $x_0,x_1 \in V$, we compute \begin{align*} (1-t)y_0 + t y_1 = (1-t)T(x_0) + tT(x_1). \end{align*} Therefore \begin{align*} (1-t)y_0 + t y_1 = T((1-t)x_0 + t x_1). \end{align*} By the definition of $x_t$, this gives \begin{align*} (1-t)y_0 + t y_1 = T(x_t). \end{align*} Since $x_t \in C$, we have $T(x_t) \in T(C)$. Hence \begin{align*} (1-t)y_0 + t y_1 \in T(C). \end{align*} Because $y_0,y_1 \in T(C)$ and $t \in [0,1]$ were arbitrary, $T(C)$ is convex. [/step] [step:Pull the convexity test back through $T$] Assume $D \subset W$ is convex. To prove that $T^{-1}(D)$ is convex, let $x_0,x_1 \in T^{-1}(D)$ and let $t \in [0,1]$. Define \begin{align*} x_t := (1-t)x_0 + t x_1. \end{align*} By the definition of preimage, $T(x_0) \in D$ and $T(x_1) \in D$. Since $D$ is convex, the point \begin{align*} (1-t)T(x_0) + tT(x_1) \end{align*} belongs to $D$. [guided] We want to prove that $T^{-1}(D)$ is convex, so we must start with two arbitrary points of $T^{-1}(D)$ and show that their convex combinations remain in $T^{-1}(D)$. Let $x_0,x_1 \in T^{-1}(D)$ and let $t \in [0,1]$. Define \begin{align*} x_t := (1-t)x_0 + t x_1. \end{align*} The target conclusion is $x_t \in T^{-1}(D)$, which by definition means $T(x_t) \in D$. Because $x_0,x_1 \in T^{-1}(D)$, their images satisfy $T(x_0) \in D$ and $T(x_1) \in D$. Now we use convexity of $D$: since $D$ contains both $T(x_0)$ and $T(x_1)$, it also contains their convex combination \begin{align*} (1-t)T(x_0) + tT(x_1). \end{align*} The reason affinity is relevant is that this convex combination in $W$ is exactly the image under $T$ of the convex combination in $V$. [/guided] [/step] [step:Conclude that the preimage is convex] By affinity of $T$, \begin{align*} T(x_t) = T((1-t)x_0 + t x_1). \end{align*} Therefore \begin{align*} T(x_t) = (1-t)T(x_0) + tT(x_1). \end{align*} The right-hand side belongs to $D$, so $T(x_t) \in D$. By the definition of preimage, $x_t \in T^{-1}(D)$. Since $x_0,x_1 \in T^{-1}(D)$ and $t \in [0,1]$ were arbitrary, $T^{-1}(D)$ is convex. This proves both assertions. [/step]