Exercise 3.E.3

Answers

Proof. Let $V = U_{1} = U_{2} = 𝔽^{\infty}$ . Note that $U_{1} + U_{2} = 𝔽^{\infty}$ . Define $T : 𝔽^{\infty} \times 𝔽^{\infty} \to 𝔽^{\infty}$ by

T ((x_{1}, x_{2}, x_{3}, \dots), (y_{1}, y_{2}, y_{3}, \dots)) = (x_{1}, y_{1}, x_{2}, y_{2}, x_{3}, y_{3}, \dots) .

Thus $T$ is obviously surjective. To prove injectivity, let

((x_{1}, x_{2}, x_{3}, \dots), (y_{1}, y_{2}, y_{3}, \dots)), ((z_{1}, z_{2}, z_{3}, \dots), (w_{1}, w_{2}, w_{3}, \dots)) \in 𝔽^{\infty} \times 𝔽^{\infty}

such that

T ((x_{1}, x_{2}, x_{3}, \dots), (y_{1}, y_{2}, y_{3}, \dots)) = T ((z_{1}, z_{2}, z_{3}, \dots), (w_{1}, w_{2}, w_{3}, \dots)) .

This implies that

(x_{1}, y_{1}, x_{2}, y_{2}, x_{3}, y_{3}, \dots) = (z_{1}, w_{1}, z_{2}, w_{2}, z_{3}, w_{3}) .

Therefore $x_{j} = z_{j}$ and $y_{j} = w_{j}$ for each $j$ . Hence

((x_{1}, x_{2}, x_{3}, \dots), (y_{1}, y_{2}, y_{3}, \dots)) = ((z_{1}, z_{2}, z_{3}, \dots), (w_{1}, w_{2}, w_{3}, \dots))

and so $T$ is injective. Thus $T$ is an isomorphism, as desired. □

celiopassos

2017-10-06 00:00