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Summary:

I have read definitions that V/U = {v + U : v ∈ V} . U is a subspace of V. But v + U is also defined as the set {v + u : u ∈ U}. So V/U is a set of sets is this the correct understanding of this?
A linear map ##\pi : V \rightarrow V/U## defined by ##\pi(v) = v + U## . So here ##\pi## takes the vector v as input and returns the set {v + u : u ∈ U} ? How is even linear independence, null space and such defined in a situation where you have a vector as input and then a set as output ?
Hi! I want to check if i have understood concepts regarding the quotient U/V correctly or not.
I have read definitions that ##V/U = \{v + U : v ∈ V\}## . U is a subspace of V. But v + U is also defined as the set ##\{v + u : u ∈ U\}##. So V/U is a set of sets is this the correct understanding of this?
A linear map ##\pi : V \rightarrow V/U## defined by ##\pi(v) = v + U## . So here ##\pi## takes the vector v as input and returns the set ##{v + u : u ∈ U}## ? How is even linear independence, null space and such defined in a situation where you have a vector as input and then a set as output ?
For example i also found the theorem that says: ##dim(V/U) = dim(V)  dim(U)## which they said holds because ##dim(V) = dim(null(\pi)) + dim(range(\pi)) = dim(U) + dim(V/U)##
I assume this must mean that ##dim(null(\pi)) = dim(U)## , why is that?
Thanks in advance!
I have read definitions that ##V/U = \{v + U : v ∈ V\}## . U is a subspace of V. But v + U is also defined as the set ##\{v + u : u ∈ U\}##. So V/U is a set of sets is this the correct understanding of this?
A linear map ##\pi : V \rightarrow V/U## defined by ##\pi(v) = v + U## . So here ##\pi## takes the vector v as input and returns the set ##{v + u : u ∈ U}## ? How is even linear independence, null space and such defined in a situation where you have a vector as input and then a set as output ?
For example i also found the theorem that says: ##dim(V/U) = dim(V)  dim(U)## which they said holds because ##dim(V) = dim(null(\pi)) + dim(range(\pi)) = dim(U) + dim(V/U)##
I assume this must mean that ##dim(null(\pi)) = dim(U)## , why is that?
Thanks in advance!