Let $A$ be an $R$-algebra for a unital commutative ring $R$, such that $a^2 = 0$ for all $a \in A$ and $\varphi:M \to A$ be an $R$-module homomorphism. We want to prove that it follows that there exists a unique $R$-algebra homomorphism $\phi:\bigwedge(M) \to A$ such that $\phi|_{M} = \varphi$.
Now, here is my thinking.
We know that $$\bigwedge(M) = \mathcal{T}(M)/\mathcal{A}(M)$$ where $\mathcal{T}(M)$ is the tensor-algebra and usually one defines $$\mathcal{A}(M) := \langle m_1 \otimes \cdots \otimes m_n \;|\; m_i \in M, \text{and} \ \exists i,j \in \{1,\ldots,k\} \ \text{such that} \ i \neq j, \text{but} \ m_i = m_j\rangle.$$
But here, I believe we want to say that $$\mathcal{A}(M) := \langle m \otimes m ;\ | \; m \in M \rangle.$$ I believe this definition and the one above is equivalent.
Now, let´s define a map $$\psi:M^n \to A$$ explicitly as $$(m_1,\ldots,m_n) \mapsto \varphi(m_1) \cdots \varphi(m_n)$$ and the injection $$\iota:M^n \to M^{\otimes n}.$$
Now, from this, we get (from another theorem, although I don´t understand it fully) a uniquely defined $R$-algebra homomorphism $$\phi':\mathcal{T}(M) \to A$$ such that $$\phi'|_{M} = \varphi$$ and so that (I believe) $$m_1 \otimes \cdots \otimes m_k \overset{\phi'}{\mapsto} \varphi(m_1) \cdots \varphi(m_k)$$
Now, we note that we naturally have a surjective $R$-module homomorphism $$q:\mathcal{T}(M) \twoheadrightarrow \mathcal{T}(M)/\mathcal{A}(M)$$
We notice that all elements $m' \in \mathcal{A}(M) \subset \mathcal{T}(M)$ get´s sent to $0$ under $\phi'$, since if we have $i \neq i+1$ but $m_i = m_{i+1}$ in $m' = m_1 \otimes \cdots \otimes m_i \otimes m_{i+1} \otimes \cdots \otimes m_k$ for some $k \in \mathbb{N}_{>0}$, where I have used the second definition of $\mathcal{A}(M)$ I defined earlier. Then $$\phi'(m') = \phi'(m_1 \otimes \cdots \otimes m_i \otimes \cdots \otimes m_j \otimes \cdots \otimes m_k)$$$$ = \varphi(m_1) \cdots \underbrace{\varphi(m_i) \varphi(m_j)}_{ = 0} \cdots \varphi(m_k) = 0.$$
Now, the first isomorphism theorem gives us a map $\bar{f}:\bigwedge(M) \to A$ such that $$\bar{f}(m_1 \wedge \cdots \wedge m_k) = \phi'(m_1 \otimes \cdots \otimes m_k)$$ and since $\phi'$ is unique (by universal property), I believe we can conclude that it follows that $\bar{f}$ is unique.
Hence $\bar{f} = \phi$. Is my reasoning correct?