Tensor, Symmetric, and Exterior Subalgebras (11.5.14)

Dummit and Foote Abstract Algebra, section 11.5, exercise 14:

Prove that when the $R$-module $M$ is a direct summand of $N$ by $φ$, then $\mathcal{T}(M)$, $\mathcal{S}(M)$, and $\bigwedge M$ are $R$-subalgebras of $\mathcal{T}(N)$, $\mathcal{S}(N)$, and $\bigwedge N$, respectively.

Proof: For each of the above algebras from $M$, we have induced algebra homomorphisms $$Φ(m_1 ⊗ ... ⊗ m_k)=φ(m_1) ⊗ ... ⊗ φ(m_k)$$ $$Φ(m_1 ⊗ ... ⊗ m_k \text{ mod }\mathcal{A}(M)) = φ(m_1) ⊗ ... ⊗ φ(m_k) \text{ mod }A(N)$$ $$Φ(m_1 ∧ ... ∧ m_k) = φ(m_1) ∧ ... ∧ φ(m_k)$$ As well, we have a natural homomorphism $φ' : N → M$ by collapsing onto the $M$ component which induces homomorphisms of algebras by $Φ'$. Since we see $φ'∘φ=1$, likewise we have $Φ'∘Φ=1$ so that $Φ$ is injective.$~\square$