I am a beginner in Category Theory so please excuse me if this is a trivial question.
Let $\mathbf{FSet}$ denote the category of finite sets. The product functor $X\times -:\mathbf{FSet}\to \mathbf{FSet}$ has a right adjoint for every finite set $X$.
My question is, does it also have a left adjoint?
Thanks!
On
No, it doesn't unless $X$ is a singleton. The very first condition to check for a functor to have a left adjoint is that it should preserve limits (such as products, equalizers...). But clearly in general, if $X$ has at least two element, $$X \times (Y \times Z) \not\cong (X \times Y) \times (X \times Z),$$ and so the functor doesn't preserve products, thus it doesn't have a left adjoint.
If $X = \varnothing$ we do have the equality above, but then $\varnothing \times \{*\} = \varnothing \neq \{*\}$ and the functor does not preserve the terminal object. So again it doesn't have a left adjoint.
However if $X = \{*\}$ itself then your functor is the identity functor, which indeed has a left adjoint: itself.
On
In this category, I find counting to be effective at disproving conjectures.
If it did have a left adjoint $F$, then there would be a bijection
$$ \hom(FA, B) \cong \hom(A, X \times B) $$
so we would have
$$ b^{|FA|} = (|X| b)^{|A|} $$
for every finite cardinal $b$. Supposing $|X| > 0$, we would need
$$ |FA| = |A| (1 + \log_b |X|) $$
for all positive integers $b$. Clearly this cannot happen if $|X| > 1$.
Good question. No, it does not have a left adjoint. One of the most important properties of adjoints is that right adjoints commute with limits (including products and the terminal object), while left adjoints commute with colimits (including coproducts and the initial object). For example, since $\emptyset$ is an initial object and $X \times -$ is a left adjoint, $X \times \emptyset \cong \emptyset$.
On the other hand, if $X \times -$ were a right adjoint (that is, if it had a left adjoint), it would have to commute with limits, including the terminal object. In $\operatorname{FSet}$, the terminal object is $1$, a singleton set. However, in general $X \times 1 \ncong 1$. So $X \times -$ does not have a left adjoint (provided $X \ncong 1$).