Question about Natural Isomorphism in Definition of Tangent Spaces

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I am a new learner in manifolds and have several questions about proof of the following Lemma:

Lemma:

$M_{m}$ is tangent space to a manifold $M$ at point $m$, and $F_{m}$ be the set of germs vanishing at $m$, then $M_{m}$ is naturally isomorphic with $(F_{m}/F_{m}^{2})^{*}$

Part of the proof:

if $l\in$ $(F_{m}/F_{m}^{2})^{*}$, we define a tangent vector $v_{l}$ at m by setting

$v_{l}$=$l$({f-f(m)}) for f$\in$ $G_{m}$,

where f(m) denotes the germ of the function with the constant value f(m), $G_{m}$ being the set of germs at m, {}denoting the cosets in $(F_{m}/F_{m}^{2})^{*}$

It's clear that $v_{l}$ is linear and a derivation(process skimped).

Thus we obtain mappings of $M_{m}$ into $(F_{m}/F_{m}^{2})^{*}$ and vice versa.

The following are my questions:

(1)The quotient group $F_{m}/F_{m}^{2}$ is defined via plus rather than multiplication, is that correct?

(2)$l\in$ $(F_{m}/F_{m}^{2})^{*}$, that means $l$ maps a germ in $(F_{m}/F_{m}^{2})^{*}$ to a real number, so does $v_{l}$=$l$({f-f(m)}) mean that all the cosets of f-f(m) are mapped to the same number?

(3)The process mentioned about gives mappings of $(F_{m}/F_{m}^{2})^{*}$ to $M_{m}$, rather than $M_{m}$ into $(F_{m}/F_{m}^{2})^{*}$ as the author claimed.

Is my understanding correct? If not, would anyone help understand the proof?

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(1) Yes. $F_m$ is a vector space under pointwise addition of representing functions (but also an $\Bbb R$-algebra) and we want to obtain a vector space, $M_m$. Of course the definition of $F_m^2$ uses the algebra structure (i.e., pointwise multiplication of representing functions). Note that for a general algebra $A$, it need nor be the case that $A^2$ is a vector subspace, but for $F_m$ this works.

(2) If $\mathbf f_1$ and $\mathbf f_2$ differ by a constant, then $\mathbf f_1-\mathbf f_1(m)=\mathbf f_2-\mathbf f_2(m)$ and hence $v_l(\mathbf f_1)=v_l(\mathbf f_2)$. But even if $\mathbf f_1,\mathbf f_2$ additionally differ by a product of germs vanishing at $m$, we have $v_l(\mathbf f_1)=v_l(\mathbf f_2)$ because $l$ evaluates on elements of $F_m/F_m^2$, not on elements of $F_m$ themselves.

(3) Yes, we obtain a linear map $(F_m/F_m^2)^*\to M_m$. An argument that this is an isomorphism is needed (by either specifying the inverse map or showing injectivity and surjectivity).