Pushout as a functor on an exact sequence I'm having some problems with the last question in exercise 2.6.4 from Weibel's "An introduction to Homological Algebra". 
The exercise asks to show that pushout is not an exact functor in Ab (Abelian categories).  I think that this shall be done by finding a counterexample but the problem arise because in the book isn't explained how the pushout should work as a functor on an  exact sequence (and I'm having problems finding this definition on books an on the web too). 
It would be welcome if someone could explains it and even more a counterexample. Thanks 
 A: The domain of the functor is to be understood as the category of $\bullet\leftarrow\bullet\rightarrow$-shaped diagrams with values in the given abelian category. This is again an abelian category, as is any diagram category over an abelian category; moreover, exactness of a sequence of such diagrams translates to exactness in each component separately. You should definitely check this if you haven't already done so.
Concretely, this means that the question asks whether for any commutative diagram in an abelian category
$$\begin{array}{lllll} 0 & \to & X_0 & \to & X_1 & \to & X_2 & \to & 0\\ &&\uparrow && \uparrow && \uparrow \\ 0 & \to & A_0 & \to & A_1 & \to & A_2 & \to & 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 & \to & Y_0 & \to & Y_1 & \to & Y_2 & \to & 0\end{array}$$
in which all three rows are exact, the induced sequence $$0\to\text{Pushout}(X_0\leftarrow A_0\rightarrow Y_0)\to\text{Pushout}(X_1\leftarrow A_1\rightarrow Y_1)\to \text{Pushout}(X_2\leftarrow A_2\rightarrow Y_2)\to 0$$ 
is exact, too. 
To understand this why this is not the case, try to find an example by restricting to the subcategory of diagrams of the form $0\leftarrow \bullet\to\bullet$ - this is an abelian subcategory of the category under consideration, with exact inclusion functor (by the component-wise nature of exactness mentioned above). Moreover, when restricted to this subcategory, the pushout functor transforms to the cokernel functor (check this!), and you might want to have a look at the Snake Lemma to see why and how this fails to be exact.
