I'm not completely sure if this bunch of questions is the appropriate Level of MO. However at the same time I think that it is at least slightly above the level of stackex. ...

The tensor algebra functor $T: \mathbf{Vec} \to \mathbf{Alg}$ that assigns the Tensor Algebra to any vector space is the left adjoint to the forgetful functor $V:\mathbf{Alg} \to \mathbf{Vec}$ that forgets the algebraic struture and just keeps the linear structure. This is reflected in the universal property of the tensor algebra. That is:

Any linear map $f$ from a vector space $V$ to an algebra $A$ can be uniquely extended to an algebra homomorphism $\bar{f}$ from $T(V)$ to $A$ such that $\bar{f}\circ i = f$ holds for the inclurion $i : V \to T(V)$.

Moreover since the right adjoint is the forgetful functor the tensor algebra is the free algebra over the vector space.

So here is the first question:

1.) The dual situation: On the underlying vector space of the tensor algebra there is more than one coalgebra structure. For example there is the shuffle coproduct and the deconcatenation coproduct. Is one of them (co)universal and moreover cofree? I.e. is the functor $\bar{T}: \mathbf{Vec} \to \mathbf{CAlg}$, that assigns to a vector space the vector space underlying the tensor algebra together with one of those coproducts, a right adjoint to the forgetful functor $V: \mathbf{CAlg}\to \mathbf{Vec}$ that forgets the coalgebraic structure?

Now the exterior algebra as well as the symmetric algebra are defined as quotients of the tensor algebra and as the tensor algebra they have a similar universal property.

2.) Is the exterior/symmetric algebra left adjoint to something? From the universal property they should be left adjoints to something.

3.) Is it a general property of universal free algebras that their quotients are universal algebras?

4.) If 3.) is true what is the dual picture for (co)universal cofree coalgebras? Do they have the property that their sub coalgebras are still (co)universal coalgebras?

I'm not completely sure if this bunch of questions is the appropriate Level of MO. However at the same time I think that it is at least slightly above the level of stackex. ...

The tensor algebra functor $T: \mathbf{Vec} \to \mathbf{Alg}$ that assigns the Tensor Algebra to any vector space is the left adjoint to the forgetful functor $V:\mathbf{Alg} \to \mathbf{Vec}$ that forgets the algebraic struture and just keeps the linear structure. This is reflected in the universal property of the tensor algebra. That is:

Any linear map $f$ from a vector space $V$ to an algebra $A$ can be uniquely extended to an algebra homomorphism $\bar{f}$ from $T(V)$ to $A$ such that $\bar{f}\circ i = f$ holds for the inclurion $i : V \to T(V)$.

Moreover since the right adjoint is the forgetful functor the tensor algebra is the free algebra over the vector space.

So here is the first question:

1.) The dual situation: On the underlying vector space of the tensor algebra there is more than one coalgebra structure. For example there is the shuffle coproduct and the deconcatenation coproduct. Is one of them (co)universal and moreover cofree? I.e. is the functor $\bar{T}: \mathbf{Vec} \to \mathbf{CAlg}$, that assigns to a vector space the vector space underlying the tensor algebra together with one of those coproducts, a right adjoint to the forgetful functor $V: \mathbf{CAlg}\to \mathbf{Vec}$ that forgets the coalgebraic structure?

Now the exterior algebra as well as the symmetric algebra are defined as quotients of the tensor algebra and as the tensor algebra they have a similar universal property.

2.) What are the right adjoints to the exterior/symmetric algebra? From the universal property they should be left adjoints to something.

3.) Is it a general property of universal free algebras that their quotients are universal algebras?

4.) If 3.) is true what is the dual picture for (co)universal cofree coalgebras? Do they have the property that their sub coalgebras are still (co)universal coalgebras?