Must two morphisms into an algebraic space which agree on closed points be the same? First I apologize if this is elementary. I have just started looking at the basics of stacks and algebraic spaces so my understanding is lacking.
Let's work over an algebraically closed field $k$. Suppose I have an algebraic space $\mathcal{A}$ and two morphisms $f_i:S\rightarrow\mathcal{A}$ such that for every closed point $\operatorname{Spec}(k)\rightarrow S$ the induced maps $f_i|_{\operatorname{Spec}(k)}\rightarrow \mathcal{A}$ agree. Suppose $S$ is a reduced scheme with $\operatorname{Spec}(k)$ points which are dense in $S$. Must $f_1=f_2$? If not is there a condition we can impose on the algebraic space so this holds?
My intuition says since an algebraic space is fibered over sets instead of groupoids this should hold.
Thanks

It was difficult to capture the essence of what I was looking at, so I apologize that I never nailed down the correct hypotheses. This exercise from Hartshorne is what I was looking for an analog of in algebraic spaces:
II.4.2: Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms which agree on an open dense subset of $X$. Then $f=g$.
