On non-representability of certain hom schemes Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(\mathbb{A}^1_k,\mathbb{A}^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(\mathbb{A}^1,\mathbb{A}^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $\mathbb{A}^1_S\to \mathbb{A}^1_S$ of $S$-schemes.)
Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.


Let X be a (positive-dimensional) variety and let $f:\mathbb{A}^1_k\to X$ be a finite morphism. How does one show that the hom-functor $\mathrm{Hom}_k(\mathbb{A}^1_k,X)$  is not representable by an algebraic space?


What have I tried? Well, there is a natural morphism Isom$(\mathbb{A}^1,\mathbb{A}^1) \to \mathrm{Hom}(\mathbb{A}^1_k,X)$ which sends $g$ to $f\circ g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?
 A: Welcome, new contributor.  Let $k$ be a field.  Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $\mathcal{O}_Y(Y)$ is a $k$-vector space of infinite dimension.  For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $\text{Spec}(k)$.  The Hom functor $\text{Hom}_{k-\text{Sch}}(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving.  Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type.  The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.  
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure.  Thus, without loss of generality, assume that $k$ is algebraically closed.  Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $\Omega_{X/k,x}\otimes_{\mathcal{O}_{X,x}} k = \mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2$ has positive dimension as a $k$-vector space.  
Denote by  $$c_x:Y\to X$$ the constant $k$-morphism with image $x$.  Then the Zariski tangent space to the Hom functor at $c_x$ equals $$\text{Hom}_{\mathcal{O}_Y}(c_x^*\Omega_{X/k},\mathcal{O}_Y) = \mathcal{O}_Y \otimes_k \left(\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\right).$$  This is a $k$-vector space of infinite dimension.  If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space.  Thus, the Hom scheme is not representable.
