Is "reverse homology" $\ker g \subset \text{im} f$ possible? $\forall$ has $\exists$ as a "left adjoint" in Topos Theory.
Let $M \xrightarrow{f} M' \xrightarrow{g} M''$  be  sequence of $A$-modules and $A$-module homomorphisms.  Now suppose we don't have $gf = 0$ or in other words no ability to compute traditional Homology at $M'$, but instead we have sort of an adjoint opposite:
$$
\forall x \in \ker g, \exists y \in M : f(y) = x
$$
which leads to $gf(y) = 0$.  In other words, we're dealing with instead of traditional homology as $\text{im}f \subset \ker g$, the opposite inclusion $\ker g \subset \text{im} f$.
There can be long sequences of these types of "reverse complexes" in practice, since any sequence of surjections and any sequence of injections is of this type, but it of course probably occurs  outside of those as well.
Do any techniques from traditional (forward) homology carry over to this "reverse homology"?
Note that a sequence is exact in this "reverse homology" if and only if it is exact in homology.
Since we could define this reverse homology, then we could define "negative dimension".  So in other words perhaps have mixed sequences of forward and reverse homologies, where exactness has dimension zero, wherever dimension is defined such as on vector spaces.
 A: We define a reverse chain complex $X_{\bullet}$ of $A$-modules and $A$-module homomorphisms $d_{i} : X_i \to X_{i-1}$ to be a usual chain complex of $A$-modules except the condition $\operatorname{im} d_i \subset \ker d_{i-1}$ gets reversed, that is: $\ker d_{i-1} \subset \operatorname{im} d_i$.
On any reverse chain complex $X_{\bullet}$ with differential $d_{i}$, we define   the $i$th reverse homology module by $H_i(X) := \dfrac{\operatorname{im}d_{i+1}}{\ker d_i}$.  Just as in usual homology we define the module of $i$-cycles by $Z_i(X) := \ker d_i$ and the module of $i$-boundaries by $B_i(X) := \operatorname{im} d_{i+1}$.  If $\alpha : X_{\bullet} \to Y_{\bullet}$ is a morphism of reverse chain complexes, then we get an induced canonical homomorphism $H_i(\alpha) : H_i(X) \to H_i(Y)$ (reading from Lang's Algebra).
Proof. The same as in usual chain complexes, $\alpha$ takes $i$-cycles of $X$ into $i$-cycles of $Y$ and $i$-boundaries of $X$ into $i$-boundaries of $Y$.  So that if $\overline{x} \in H_i(X)$, then $\overline{x} = x + Z_i(X)$ for some $x \in \operatorname{im} d_{i+1}$.  Define $H_i(\alpha)(\overline{x}) := \alpha(x) + Z_i(Y)$.  Then $H_i(\alpha)$ is well-defined, for if $\overline{x} = \overline{y}$, then $x - y \in Z_i(X)$ which means $\alpha(x) - \alpha(y) \in Z_i(Y)$ by the first sentence.  We have that $H_i(\alpha)$ is an $A$-module homomorphism since $$H_i(\alpha)(\overline{x} + \overline{y}) = H_i(\alpha)(\overline{x + y}) = \alpha(x + y) + Z_i(Y) = \alpha(x) + \alpha(y) + Z_i(Y) = (\alpha(x) + Z_i(Y)) + (\alpha(y) + Z_i(Y)) = H_i(\alpha)(\overline{x}) + H_i(\alpha)(\overline{y})$$
And finally for $\gamma \in A$, we have $H_i(\alpha)(\gamma \overline{x}) = H_i(\alpha)(\overline{\gamma x}) = \alpha(\gamma x) + Z_i(Y) = \gamma \alpha(x) + Z_i(Y) = \gamma H_i(\alpha)(x)$.

Furthermore, if $X_{\bullet}$ is a reverse chain complex of $A$-modules and $d_i$ its differential, then a sequence of $A$-modules $H_i(X)$ and maps $H_i(d)$ gets induced.
Proof.  Let $\overline{x} = x + Z_i(X)$ and define $H_i(d)(\overline{x}) = d_i(x) + Z_{i-1}(X)$.  Then if $\overline{x} = \overline{y}$, we have $\overline{x - y} = 0$ or $x - y \in Z_i(X)$ so that $d_i(x-y) = 0$ and $d_{i-1}\circ d_i(x - y) = 0$ and so $d_i(x) - d_i(y) \in Z_{i-1}(X)$ and we're done.  We leave it to the reader to prove that $H_i(d)$ so defined is an $A$-module homomorphism.
Thus, we have proven that $H_i(X)$ acts somewhat like the $i$th homology module from  General Homology Theory.
I don't yet know if there is a functor $H$ from the category of $A$-module reverse chain complexes to the category of graded $A$-modules.  Can someone who is more proficient at HA comment or answer?
