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Let $(V,+,\cdot)$ be a vector space over field $K$ and $(T(V),+_T,\cdot_T,\times_T)$ be it's tensor algebra. Where: $$T(V)=\bigoplus\limits_{k=0}^{\infty}T^kV,\quad T(V)\ni v=(v_0,v_1,v_2,\dots)$$ (Where $v_0 \in K$, $v_1 \in V$, $v_2 \in V\otimes V$, $\dots $ and $v_i$=0 for cofinite number.)
And multiplication defined by: $$\forall v,w \in T(V): (v \times_Tw)=(v_0\otimes w_0,v_1 \otimes w_0+v_0\otimes w_1,v_0\otimes w_2+v_2 \otimes v_0+v_1\otimes w_1,\dots)$$ The exterior algebra $\Lambda(V)=\frac{T(V)}{I}$ is a quotient algebra of $T(V)$ by the twosided ideal $I$ generated by the set: $$ X=\{(0_R,0_V,x_i \otimes x_i,0_{V^\otimes 3},\dots)|x_i\in V\}\subset T(V)\\ I=\{\sum\limits_{i}((p_{0i},p_{1i},p_{2i},\dots)\times_T(0_R,0_V,x_i\otimes x_i,0_{V^\otimes 3},\dots)\times_T(s_{0i},s_{1i},s_{2i},\dots))|x_i \in V;\\p,s\in T(V)\}$$ Congruence generated by $I$: $\forall a,b \in T(V): a\sim b:\Leftrightarrow (a-b)\in I$
So then: $\Lambda(V)=\{[v]_\sim :v\in T(V)\}=\{\{w \in T(V)|v-w\in I\}|v\in(T)\}$
And :
$$\wedge:\Lambda(V)\times \Lambda(V) \to \Lambda(V) \\ \forall \alpha,\beta\in \Lambda(V): \alpha \wedge \beta = [a]_\sim\wedge [b]_\sim=[(a_0,a_1,a_2,\dots)]_\sim \wedge [(b_0,b_1,b_2,\dots)]_\sim := [a\times_T b]_\sim$$ Then we define k-th exterior power $\Lambda^k(V)$ as vector subspace, such that: $$\Lambda(V)\supseteq\Lambda^k(V):=span(X)\\ X:=\{[(0,x_1,0,\dots)]_\sim\wedge \dots \wedge[(0,x_k,0,\dots)]_\sim|x_i \in V\}$$ Now if we choose $V=T^*_p M$, for point $p$ of some smooth manifold $M$. Then smooth differential $k$-forms at point $p$ are the elements of $\Lambda^k(T^*_pM)$

Question: It's often said that differential forms at point are some sort of tensors, but how can we formally speak of them that way if in fact they are just linear combinations of equivalence classes of some sequences from tensor algebra?

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    $\begingroup$ there is a natural injective map $\bigwedge V \to TV$ defined by $v_1\wedge\dots\wedge v_k\mapsto \sum_\sigma (\operatorname{sgn}\sigma)v_{\sigma(1)}\otimes\dots \otimes v_{\sigma(n)}$; this is what is used to see differential forms as tensors. In fact, one often defines $k$-forms as skew-symmetric $k$-linear maps, which makes them to tensors (sections of ${T^*}^{\otimes k}M$ are $k$-linear maps). $\endgroup$ – user8268 Apr 22 '17 at 18:09
  • $\begingroup$ @user8268 How do you define that injection if elements of $\Lambda (V)$ are equivalence classes and elements of $T(V)$ are sequences? $\endgroup$ – Sergey Dylda Apr 22 '17 at 18:13
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    $\begingroup$ what I wrote is an injection $\bigwedge^k V\to T^k V$ $\endgroup$ – user8268 Apr 22 '17 at 18:35
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Viewing the exterior power this way can be very cumbersome. Tensors in $V^{\otimes n}$ are best viewed via their universal property: any multilinear map $f: V^n \to U$ uniquely factors through the tensor product, and so specifying where a linear map sends a tensor $v_1 \otimes \cdots \otimes v_n$ is the same as specifying where a bilinear map sends the tuple $(v_1, \ldots, v_n)$. Put another way: if all you can work with are linear maps, but you need to express a bilinear map out of $V^n$, use a linear map out of $V^{\otimes n}$ instead.

Tensors in an exterior power are analogous. Call a multilinear map $f: V^n \to V$ alternating if $f(v_1, \ldots, v_n) = 0$ whenever any $v_i = v_j$ for $i \neq j$. The $n$th exterior power $\bigwedge^n V$ has the property that any multilinear alternating map $f: V^n \to U$ uniquely factors through $\bigwedge^n V$, and therefore may be viewed as a kind of "alternating tensor".

The notes here define the exterior power as a quotient of $V^{\otimes n}$ by a certain ideal, and then quickly show the universal property, and the rest of the notes just use this universal property. I think understanding these notes will mostly answer your question.

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