Assume Y, T, f, and \alpha given. Then there is the diagonal map {T \rightarrow T \times T} as always (which maps every element t to {\langle t,t \rangle}), so we can form the three-fold composite {g: T \rightarrow Y}. This new map by its construction satisfies g(t) = \alpha(f(t,t)) for every point t of T. We have assumed that every map {T \rightarrow Y} is named as {f(-,x)} for some point {\mathbf{1} \xrightarrow{x} T}, and g is such a map {T \rightarrow Y}. So let {x = t_0} be a parameter value corresponding to our g, i.e. {g = f(-,t_0)}, so that g(t) = f(t,t_0) for all t. Taking the special case t = t_0, we have g(t_0) = f(t_0,t_0) which by the definition of g says \alpha(f(t_0,t_0)) = f(t_0,t_0) or, in other words, that {y_0 = f(t_0,t_0)} defines a point of Y which is fixed by \alpha: \alpha(y_0) = y_0

Use \alpha and the diagonal as above to make f itself produce an example g which f leaves out.