Existence of a function from a quotient

Question

On "Introduction to Smooth Manifolds" by John M. Lee, at page 309 we're talking about multilinear algebra. There's a proposition on the Characteristic Property of the Tensor Product Space. In the proof he says: "The fact that $A$ is multilinear means precisely that the subspace $R$ is contained in the kernel of $\bar{A}$" and so "Therefore, $\bar{A}$ descends to a linear map $\tilde{A}$". How $R \subseteq ker\bar{A}$ justifies the existence of the map $\tilde{A}$. Also speaking at set level is it right to say that (in this particular case) $${X\over ker \bar{A} }\subseteq {X\over R} $$ if $R \subseteq ker \bar{A}$, thank you very much.

Answer 1

I'm sure this is a much more general fact about "quotients", but I will keep to the setting of linear algebra.

Let $A:V\to W$ be a linear map between two vector spaces. Also let $R\subset V$ be a subspace, so that we can form the quotient vector space $V/R = \{v+R:v\in V\}$, where $v+R = v+r+R$ whenever $r\in R$.

If $\ker A$ contains $R$, then we can define $$\tilde A(v+R):= A(v).$$ This is well defined because if $r\in R$, then $$\tilde A((v+r)+R) = A(v+r) = A(v) +A(r) = A(v)= \tilde A(v+R).$$

I'm not sure what you mean by the "quotient of sets" you have written there. One fact which may help your intuition is that if $\ker A$ contains the subspace $R$ (so $\ker A$ is "bigger"), then we will have $$\dim (V/\ker A) \leq \dim (V/R)$$ ("dividing by a bigger thing gives a smaller thing")