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Let $V_i$ be a collection of vector spaces over field $F$ where $i=1,2,...,N$. Given the Cartesian Product $V=V_1\times V_2\times...\times V_N$ equipped with natural projections $p_i:V\to V_i$.

Let $U$ be a vector space and $T_i:U\to V_i$ linear transformations. Prove there exists a unique $T$ such that $p_i\circ T=T_i$.

My thought: I don't get how exactly $p_i$ is defined but I am guessing it maps $(x_1,x_2,...,x_N)$ to $x_i$. In this case, I have shown that $p_i$ is a linear transformation and $ker\ p_i=\{0\}$. Could someone explain what natural projection means in this context? I can kind of see the parallel between this and the definition of dual basis but don't know how to approach. Should I begin by constructing a basis for $U$ and try to work something out using the fact that both $T_i$ and $p_i$ are linear? Finally how would you normally prove the existence and uniqueness of a map?

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    $\begingroup$ Edit your post to include your own attempt at a proof. $\endgroup$ – Mike Pierce Apr 23 '15 at 4:45
  • $\begingroup$ @NajibIdrissi sorry I forgot to say it is a cartesian product. $\endgroup$ – Zhaoman Apr 23 '15 at 9:22
  • $\begingroup$ @mapierce271 I just did. $\endgroup$ – Zhaoman Apr 23 '15 at 9:23
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(1)Define $$T: U\rightarrow \prod V_i,\ T(x)=(T_1(x),\cdots , T_n(x) )$$

Hence we have an existence.

(2) Note that any vector in $\prod V_i$ has the form $$ v=(v_1,\cdots, v_n)$$ where $v_i\in V_i$. Hence for all $i$, if $0=p_i(v)$, then $v=(0,\cdots, 0)$.

(3) If $S$ is another then $$ p_i\circ (S-T)(x)= 0$$ Hence $(S-T)(x)=0$. Hence we have uniqueness.

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    $\begingroup$ Sorry I don't understand the third step. How did you get pi∘(S−T)(x)=0? $\endgroup$ – Zhaoman Apr 25 '15 at 4:10
  • $\begingroup$ It is an assumption of $S$ : $p_i\circ S=T_i$ Hence $$ p_i\circ (S-T)=p_i\circ S - p_i\circ T = T_i-T_i=0$$ $\endgroup$ – HK Lee Apr 25 '15 at 4:22

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