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The image below is from my lecture notes for Linear analysis: Lecture notes on translation operator

Shouldn't it be $T:X\to X$? If not, I have no idea what it means, because if $z$ is in $X$ then there is no reason as to why $x + z$ should be in $Y$.

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    Could you write it in TeX? I don't get what is is written there!2012-05-07
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    @Norbert "Because $T$ is linear, you can use translations. For $x\in X$, the map $$\eqalign{X&\rightarrow Y \cr z&\mapsto x+z }$$ is continuous..."2012-05-07
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    Yeah that's what it looks like. I think it is a mistake and the easiest way to correct is it to change Y to X. The only other thing I think it could mean is: get a linear operator T, and y in Y, and map x (in X) to T(x) + y. Not sure this works. Anyhow, I think the first way was intended.2012-05-07
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    @DavidMitra, thanks!2012-05-07
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    @AdamRubinson I can see that after the definition of that map its inverse is also mentioned. From their forms it is clear that there is no need in $T$ operator.2012-05-07
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    I don't understand...2012-05-07
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    You see that and the end of this notes mentioned operator that is inverse to the first one. The inverse is $z\mapsto -x+z$, the original onei is $z\mapsto x+z$2012-05-08
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    How can you start talking about an inverse operator, when you have not establish the thing in the first place is actually an operator? I am still trying to establish what the map actually is. What is Y? Is Y different to X? If so, how can x be in X, z be in X, and x+z be in Y if X and Y are not equal spaces?2012-05-08
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    @AdamRubinson the map is the shift on vector $x$, the inverse map is the shift on opposite vector $-x$2012-05-08
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3376/discussion-between-adam-rubinson-and-norbert)2012-05-09
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    So does Y=X or not?2012-05-09

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