Injectivity and surjectivity of $\prod_{i=1}^m W_i \to \sum_{i=1}^m W_i$ given by $(w_i)_i \mapsto \sum_i w_i$

Question

Suppose $W_1,W_2,\dotsc,W_m$ are subspaces of $V$. Define a map $$ T \colon W_1 \times W_2 \times \dotsb \times W_m \to W_1 + W_2 + \dotsb + W_m $$ by $$ T(w_1,w_2,...,w_m) = w_1 + w_2 + \dotsb + w_m. $$

a) Show that $T$ is surjective.

b) Show that $T$ is injective if and only if $W_1 + W_2 + \dotsb + W_m$ is a direct sum.

I know that if $\dim( \operatorname{im}(T)) = \dim(W_1 \times W_2 \times \dotsb \times W_m)$, then $T$ is surjective and if $\dim(\ker(T))=0$, then $T$ is injective. I am not sure how to proceed with the proof.

Answer 1

• Suppose $r \in W_1 + W_2 + \ldots + W_m$, then $\exists w_i \in W_i, i \in \{ 1, \ldots, m\}$ such that $r=w_1 + \ldots + w_m$. Hence it is surjective.

• Suppose it is injective, then for every vector in $W_1+\ldots +W_m$, there is exactly one way to decompose the vector. In particular, there is exactly one way to decompose the zero vector, hence it is a direct sum.

• Suppose it is a direct sum, then there is exactly one way to decompose the zero vector. I will leave this as an exercise, try to show that it is an injection.